What is the general solution for sec^2(3θ)cos^2(3θ)=sec^2(3θ)+cos^2(3θ) ?

The general solution for sec^2(3θ)cos^2(3θ)=sec^2(3θ)+cos^2(3θ) is No Solution for θ\in\mathbb{R}

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sec^2(3θ)cos^2(3θ)=sec^2(3θ)+cos^2(3θ)

Solution

sec^{2} (3 θ) cos^{2} (3 θ) = sec^{2} (3 θ) + cos^{2} (3 θ)

Solution

No Solution for θ \in R

Solution steps

sec^{2} (3 θ) cos^{2} (3 θ) = sec^{2} (3 θ) + cos^{2} (3 θ)

Subtract

sec^{2} (3 θ) + cos^{2} (3 θ)

from both sides

sec^{2} (3 θ) cos^{2} (3 θ) - sec^{2} (3 θ) - cos^{2} (3 θ) = 0

Rewrite using trig identities

- cos^{2} (3 θ) - sec^{2} (3 θ) + cos^{2} (3 θ) sec^{2} (3 θ)

Use the basic trigonometric identity:

cos (x) = \frac{1}{sec ( x )}

= - (\frac{1}{sec ( 3 θ )})^{2} - sec^{2} (3 θ) + (\frac{1}{sec ( 3 θ )})^{2} sec^{2} (3 θ)

Simplify

- (\frac{1}{sec ( 3 θ )})^{2} - sec^{2} (3 θ) + (\frac{1}{sec ( 3 θ )})^{2} sec^{2} (3 θ) : - \frac{1}{sec ^{2} ( 3 θ )} - sec^{2} (3 θ) + 1

- (\frac{1}{sec ( 3 θ )})^{2} - sec^{2} (3 θ) + (\frac{1}{sec ( 3 θ )})^{2} sec^{2} (3 θ)

(\frac{1}{sec ( 3 θ )})^{2} = \frac{1}{sec ^{2} ( 3 θ )}

(\frac{1}{sec ( 3 θ )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{sec ^{2} ( 3 θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{sec ^{2} ( 3 θ )}

(\frac{1}{sec ( 3 θ )})^{2} sec^{2} (3 θ) = 1

(\frac{1}{sec ( 3 θ )})^{2} sec^{2} (3 θ)

(\frac{1}{sec ( 3 θ )})^{2} = \frac{1}{sec ^{2} ( 3 θ )}

(\frac{1}{sec ( 3 θ )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{sec ^{2} ( 3 θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{sec ^{2} ( 3 θ )}

= \frac{1}{sec ^{2} ( 3 θ )} sec^{2} (3 θ)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot sec ^{2} ( 3 θ )}{sec ^{2} ( 3 θ )}

Cancel the common factor:

sec^{2} (3 θ)

= 1

= - \frac{1}{sec ^{2} ( 3 θ )} - sec^{2} (3 θ) + 1

= - \frac{1}{sec ^{2} ( 3 θ )} - sec^{2} (3 θ) + 1

1 - \frac{1}{sec ^{2} ( 3 θ )} - sec^{2} (3 θ) = 0

Solve by substitution

1 - \frac{1}{sec ^{2} ( 3 θ )} - sec^{2} (3 θ) = 0

Let:

sec (3 θ) = u

1 - \frac{1}{u ^{2}} - u^{2} = 0

1 - \frac{1}{u ^{2}} - u^{2} = 0 : u = - \frac{3}{2} + \frac{1}{2} i, u = \frac{3}{2} - \frac{1}{2} i, u = \frac{3}{2} + \frac{1}{2} i, u = - \frac{3}{2} - \frac{1}{2} i

1 - \frac{1}{u ^{2}} - u^{2} = 0

Multiply both sides by

u^{2}

1 - \frac{1}{u ^{2}} - u^{2} = 0

Multiply both sides by

u^{2}

1 \cdot u^{2} - \frac{1}{u ^{2}} u^{2} - u^{2} u^{2} = 0 \cdot u^{2}

Simplify

1 \cdot u^{2} - \frac{1}{u ^{2}} u^{2} - u^{2} u^{2} = 0 \cdot u^{2}

Simplify

1 \cdot u^{2} : u^{2}

1 \cdot u^{2}

Multiply:

1 \cdot u^{2} = u^{2}

= u^{2}

Simplify

- \frac{1}{u ^{2}} u^{2} : - 1

- \frac{1}{u ^{2}} u^{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= - \frac{1 \cdot u ^{2}}{u ^{2}}

Cancel the common factor:

u^{2}

= - 1

Simplify

- u^{2} u^{2} : - u^{4}

- u^{2} u^{2}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

u^{2} u^{2} = u^{2 + 2}

= - u^{2 + 2}

Add the numbers:

2 + 2 = 4

= - u^{4}

Simplify

0 \cdot u^{2} : 0

0 \cdot u^{2}

Apply rule

0 \cdot a = 0

= 0

u^{2} - 1 - u^{4} = 0

u^{2} - 1 - u^{4} = 0

u^{2} - 1 - u^{4} = 0

Solve

u^{2} - 1 - u^{4} = 0 : u = - \frac{3}{2} + \frac{1}{2} i, u = \frac{3}{2} - \frac{1}{2} i, u = \frac{3}{2} + \frac{1}{2} i, u = - \frac{3}{2} - \frac{1}{2} i

u^{2} - 1 - u^{4} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

- u^{4} + u^{2} - 1 = 0

Rewrite the equation with

x = u^{2}

and

x^{2} = u^{4}

- x^{2} + x - 1 = 0

Solve

- x^{2} + x - 1 = 0 : x = \frac{1}{2} - i \frac{3}{2}, x = \frac{1}{2} + i \frac{3}{2}

- x^{2} + x - 1 = 0

Solve with the quadratic formula

- x^{2} + x - 1 = 0

Quadratic Equation Formula:

For

a = - 1, b = 1, c = - 1

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) ( - 1 )}{2 ( - 1 )}

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) ( - 1 )}{2 ( - 1 )}

Simplify

1^{2} - 4 (- 1) (- 1) : 3 i

1^{2} - 4 (- 1) (- 1)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) (- 1)

Apply rule

- (- a) = a

= 1 - 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 1 - 4

Subtract the numbers:

1 - 4 = - 3

= - 3

Apply radical rule:

- a = - 1 a

- 3 = - 1 3

= - 1 3

Apply imaginary number rule:

- 1 = i

= 3 i

x_{1, 2} = \frac{- 1 \pm 3 i}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 1 + 3 i}{2 ( - 1 )}, x_{2} = \frac{- 1 - 3 i}{2 ( - 1 )}

x = \frac{- 1 + 3 i}{2 ( - 1 )} : \frac{1}{2} - i \frac{3}{2}

\frac{- 1 + 3 i}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 3 i}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 + 3 i}{- 2}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{- 1 + 3 i}{2}

Rewrite

- \frac{- 1 + 3 i}{2}

in standard complex form:

\frac{1}{2} - \frac{3}{2} i

- \frac{- 1 + 3 i}{2}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{- 1 + 3 i}{2} = - (- \frac{1}{2}) - (\frac{3 i}{2})

= - (- \frac{1}{2}) - (\frac{3 i}{2})

Remove parentheses:

(a) = a, - (- a) = a

= \frac{1}{2} - \frac{3 i}{2}

= \frac{1}{2} - \frac{3}{2} i

x = \frac{- 1 - 3 i}{2 ( - 1 )} : \frac{1}{2} + i \frac{3}{2}

\frac{- 1 - 3 i}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 3 i}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 - 3 i}{- 2}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{- 1 - 3 i}{2}

Rewrite

- \frac{- 1 - 3 i}{2}

in standard complex form:

\frac{1}{2} + \frac{3}{2} i

- \frac{- 1 - 3 i}{2}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{- 1 - 3 i}{2} = - (- \frac{1}{2}) - (- \frac{3 i}{2})

= - (- \frac{1}{2}) - (- \frac{3 i}{2})

Apply rule

- (- a) = a

= \frac{1}{2} + \frac{3 i}{2}

= \frac{1}{2} + \frac{3}{2} i

The solutions to the quadratic equation are:

x = \frac{1}{2} - i \frac{3}{2}, x = \frac{1}{2} + i \frac{3}{2}

x = \frac{1}{2} - i \frac{3}{2}, x = \frac{1}{2} + i \frac{3}{2}

Substitute back

x = u^{2},

solve for

u

Solve

u^{2} = \frac{1}{2} - i \frac{3}{2} : u = - \frac{3}{2} + \frac{1}{2} i, u = \frac{3}{2} - \frac{1}{2} i

u^{2} = \frac{1}{2} - i \frac{3}{2}

Substitute

u = x + y i

(x + y i)^{2} = \frac{1}{2} - i \frac{3}{2}

Expand

(x + y i)^{2} : (x^{2} - y^{2}) + 2 i x y

(x + y i)^{2}

= (x + i y)^{2}

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = x, b = y i

= x^{2} + 2 x y i + (y i)^{2}

(y i)^{2} = - y^{2}

(y i)^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= i^{2} y^{2}

i^{2} = - 1

i^{2}

Apply imaginary number rule:

i^{2} = - 1

= - 1

= (- 1) y^{2}

Refine

= - y^{2}

= x^{2} + 2 i x y - y^{2}

Rewrite

x^{2} + 2 i x y - y^{2}

in standard complex form:

(x^{2} - y^{2}) + 2 x y i

x^{2} + 2 i x y - y^{2}

Group the real part and the imaginary part of the complex number

= (x^{2} - y^{2}) + 2 x y i

= (x^{2} - y^{2}) + 2 x y i

(x^{2} - y^{2}) + 2 i x y = \frac{1}{2} - i \frac{3}{2}

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[x^{2} - y^{2} = \frac{1}{2} 2 x y = - \frac{3}{2}]

[x^{2} - y^{2} = \frac{1}{2} 2 x y = - \frac{3}{2}] : (x = - \frac{3}{2}, x = \frac{3}{2}, y = \frac{1}{2} y = - \frac{1}{2})

[x^{2} - y^{2} = \frac{1}{2} 2 x y = - \frac{3}{2}]

Isolate

x

for

2 x y = - \frac{3}{2} : x = - \frac{3}{4 y}

2 x y = - \frac{3}{2}

Divide both sides by

2 y

2 x y = - \frac{3}{2}

Divide both sides by

2 y

\frac{2 x y}{2 y} = \frac{- \frac{3}{2}}{2 y}

Simplify

\frac{2 x y}{2 y} = \frac{- \frac{3}{2}}{2 y}

Simplify

\frac{2 x y}{2 y} : x

\frac{2 x y}{2 y}

Divide the numbers:

\frac{2}{2} = 1

= \frac{x y}{y}

Cancel the common factor:

y

= x

Simplify

\frac{- \frac{3}{2}}{2 y} : - \frac{3}{4 y}

\frac{- \frac{3}{2}}{2 y}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{\frac{3}{2}}{2 y}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

\frac{\frac{3}{2}}{2 y} = \frac{3}{2 \cdot 2 y}

= - \frac{3}{2 \cdot 2 y}

Multiply the numbers:

2 \cdot 2 = 4

= - \frac{3}{4 y}

x = - \frac{3}{4 y}

x = - \frac{3}{4 y}

x = - \frac{3}{4 y}

Plug the solutions

x = - \frac{3}{4 y}

into

x^{2} - y^{2} = \frac{1}{2}

For

x^{2} - y^{2} = \frac{1}{2}

, subsitute

x

with

- \frac{3}{4 y} : y = \frac{1}{2}, y = - \frac{1}{2}

For

x^{2} - y^{2} = \frac{1}{2}

, subsitute

x

with

- \frac{3}{4 y}

(- \frac{3}{4 y})^{2} - y^{2} = \frac{1}{2}

Solve

(- \frac{3}{4 y})^{2} - y^{2} = \frac{1}{2} : y = \frac{1}{2}, y = - \frac{1}{2}

(- \frac{3}{4 y})^{2} - y^{2} = \frac{1}{2}

Multiply by LCM

(- \frac{3}{4 y})^{2} - y^{2} = \frac{1}{2}

Simplify

(- \frac{3}{4 y})^{2} : \frac{3}{16 y ^{2}}

(- \frac{3}{4 y})^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- \frac{3}{4 y})^{2} = (\frac{3}{4 y})^{2}

= (\frac{3}{4 y})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{( 3 ) ^{2}}{( 4 y ) ^{2}}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

(4 y)^{2} = 4^{2} y^{2}

= \frac{( 3 ) ^{2}}{4 ^{2} y ^{2}}

(3)^{2} : 3

Apply radical rule:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 3^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 3

= \frac{3}{4 ^{2} y ^{2}}

4^{2} = 16

= \frac{3}{16 y ^{2}}

\frac{3}{16 y ^{2}} - y^{2} = \frac{1}{2}

Find Least Common Multiplier of

16 y^{2}, 2 : 16 y^{2}

16 y^{2}, 2

Lowest Common Multiplier (LCM)

Least Common Multiplier of

16, 2 : 16

16, 2

Least Common Multiplier (LCM)

Prime factorization of

16 : 2 \cdot 2 \cdot 2 \cdot 2

16

16

divides by

216 = 8 \cdot 2

= 2 \cdot 8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Multiply each factor the greatest number of times it occurs in either

16

2

= 2 \cdot 2 \cdot 2 \cdot 2

Multiply the numbers:

2 \cdot 2 \cdot 2 \cdot 2 = 16

= 16

Compute an expression comprised of factors that appear either in

16 y^{2}

2

= 16 y^{2}

Multiply by LCM=

16 y^{2}

\frac{3}{16 y ^{2}} \cdot 16 y^{2} - y^{2} \cdot 16 y^{2} = \frac{1}{2} \cdot 16 y^{2}

Simplify

\frac{3}{16 y ^{2}} \cdot 16 y^{2} - y^{2} \cdot 16 y^{2} = \frac{1}{2} \cdot 16 y^{2}

Simplify

\frac{3}{16 y ^{2}} \cdot 16 y^{2} : 3

\frac{3}{16 y ^{2}} \cdot 16 y^{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{3 \cdot 16 y ^{2}}{16 y ^{2}}

Cancel the common factor:

16

= \frac{3 y ^{2}}{y ^{2}}

Cancel the common factor:

y^{2}

= 3

Simplify

- y^{2} \cdot 16 y^{2} : - 16 y^{4}

- y^{2} \cdot 16 y^{2}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

y^{2} y^{2} = y^{2 + 2}

= - 16 y^{2 + 2}

Add the numbers:

2 + 2 = 4

= - 16 y^{4}

Simplify

\frac{1}{2} \cdot 16 y^{2} : 8 y^{2}

\frac{1}{2} \cdot 16 y^{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 16}{2} y^{2}

\frac{1 \cdot 16}{2} = 8

\frac{1 \cdot 16}{2}

Multiply the numbers:

1 \cdot 16 = 16

= \frac{16}{2}

Divide the numbers:

\frac{16}{2} = 8

= 8

= 8 y^{2}

3 - 16 y^{4} = 8 y^{2}

3 - 16 y^{4} = 8 y^{2}

3 - 16 y^{4} = 8 y^{2}

Solve

3 - 16 y^{4} = 8 y^{2} : y = \frac{1}{2}, y = - \frac{1}{2}

3 - 16 y^{4} = 8 y^{2}

Move

8 y^{2}

to the left side

3 - 16 y^{4} = 8 y^{2}

Subtract

8 y^{2}

from both sides

3 - 16 y^{4} - 8 y^{2} = 8 y^{2} - 8 y^{2}

Simplify

3 - 16 y^{4} - 8 y^{2} = 0

3 - 16 y^{4} - 8 y^{2} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

- 16 y^{4} - 8 y^{2} + 3 = 0

Rewrite the equation with

u = y^{2}

and

u^{2} = y^{4}

- 16 u^{2} - 8 u + 3 = 0

Solve

- 16 u^{2} - 8 u + 3 = 0 : u = - \frac{3}{4}, u = \frac{1}{4}

- 16 u^{2} - 8 u + 3 = 0

Solve with the quadratic formula

- 16 u^{2} - 8 u + 3 = 0

Quadratic Equation Formula:

For

a = - 16, b = - 8, c = 3

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 16 ) \cdot 3}{2 ( - 16 )}

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 16 ) \cdot 3}{2 ( - 16 )}

(- 8)^{2} - 4 (- 16) \cdot 3 = 16

(- 8)^{2} - 4 (- 16) \cdot 3

Apply rule

- (- a) = a

= (- 8)^{2} + 4 \cdot 16 \cdot 3

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 8)^{2} = 8^{2}

= 8^{2} + 4 \cdot 16 \cdot 3

Multiply the numbers:

4 \cdot 16 \cdot 3 = 192

= 8^{2} + 192

8^{2} = 64

= 64 + 192

Add the numbers:

64 + 192 = 256

= 256

Factor the number:

256 = 1 6^{2}

= 1 6^{2}

Apply radical rule:

1 6^{2} = 16

= 16

u_{1, 2} = \frac{- ( - 8 ) \pm 16}{2 ( - 16 )}

Separate the solutions

u_{1} = \frac{- ( - 8 ) + 16}{2 ( - 16 )}, u_{2} = \frac{- ( - 8 ) - 16}{2 ( - 16 )}

u = \frac{- ( - 8 ) + 16}{2 ( - 16 )} : - \frac{3}{4}

\frac{- ( - 8 ) + 16}{2 ( - 16 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{8 + 16}{- 2 \cdot 16}

Add the numbers:

8 + 16 = 24

= \frac{24}{- 2 \cdot 16}

Multiply the numbers:

2 \cdot 16 = 32

= \frac{24}{- 32}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{24}{32}

Cancel the common factor:

8

= - \frac{3}{4}

u = \frac{- ( - 8 ) - 16}{2 ( - 16 )} : \frac{1}{4}

\frac{- ( - 8 ) - 16}{2 ( - 16 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{8 - 16}{- 2 \cdot 16}

Subtract the numbers:

8 - 16 = - 8

= \frac{- 8}{- 2 \cdot 16}

Multiply the numbers:

2 \cdot 16 = 32

= \frac{- 8}{- 32}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{8}{32}

Cancel the common factor:

8

= \frac{1}{4}

The solutions to the quadratic equation are:

u = - \frac{3}{4}, u = \frac{1}{4}

u = - \frac{3}{4}, u = \frac{1}{4}

Substitute back

u = y^{2},

solve for

y

Solve

y^{2} = - \frac{3}{4} :

No Solution for

y \in R

y^{2} = - \frac{3}{4}

x^{2}

cannot be negative for

x \in R

No Solution for y \in R

Solve

y^{2} = \frac{1}{4} : y = \frac{1}{2}, y = - \frac{1}{2}

y^{2} = \frac{1}{4}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

y = \frac{1}{4}, y = - \frac{1}{4}

\frac{1}{4} = \frac{1}{2}

\frac{1}{4}

Apply radical rule:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= \frac{1}{4}

Apply radical rule:

1 = 1

1 = 1

= \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

a^{2} = a, a \geq 0

2^{2} = 2

= 2

= \frac{1}{2}

- \frac{1}{4} = - \frac{1}{2}

- \frac{1}{4}

Apply radical rule:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= - \frac{1}{4}

Apply radical rule:

1 = 1

1 = 1

= - \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

a^{2} = a, a \geq 0

2^{2} = 2

= 2

= - \frac{1}{2}

y = \frac{1}{2}, y = - \frac{1}{2}

The solutions are

y = \frac{1}{2}, y = - \frac{1}{2}

y = \frac{1}{2}, y = - \frac{1}{2}

Verify Solutions

Find undefined (singularity) points:

y = 0

Take the denominator(s) of

(- \frac{3}{4 y})^{2} - y^{2}

and compare to zero

Solve

4 y = 0 : y = 0

4 y = 0

Divide both sides by

4

4 y = 0

Divide both sides by

4

\frac{4 y}{4} = \frac{0}{4}

Simplify

y = 0

y = 0

The following points are undefined

y = 0

Combine undefined points with solutions:

y = \frac{1}{2}, y = - \frac{1}{2}

Plug the solutions

y = \frac{1}{2}, y = - \frac{1}{2}

into

2 x y = - \frac{3}{2}

For

2 x y = - \frac{3}{2}

, subsitute

y

with

\frac{1}{2} : x = - \frac{3}{2}

For

2 x y = - \frac{3}{2}

, subsitute

y

with

\frac{1}{2}

2 x \frac{1}{2} = - \frac{3}{2}

Solve

2 x \frac{1}{2} = - \frac{3}{2} : x = - \frac{3}{2}

2 x \frac{1}{2} = - \frac{3}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

\frac{1 \cdot 2}{2} x = - \frac{3}{2}

Cancel the common factor:

2

x \cdot 1 = - \frac{3}{2}

Multiply:

x \cdot 1 = x

x = - \frac{3}{2}

For

2 x y = - \frac{3}{2}

, subsitute

y

with

- \frac{1}{2} : x = \frac{3}{2}

For

2 x y = - \frac{3}{2}

, subsitute

y

with

- \frac{1}{2}

2 x (- \frac{1}{2}) = - \frac{3}{2}

Solve

2 x (- \frac{1}{2}) = - \frac{3}{2} : x = \frac{3}{2}

2 x (- \frac{1}{2}) = - \frac{3}{2}

Divide both sides by

2 (- \frac{1}{2})

2 x (- \frac{1}{2}) = - \frac{3}{2}

Divide both sides by

2 (- \frac{1}{2})

\frac{2 x ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )} = \frac{- \frac{3}{2}}{2 ( - \frac{1}{2} )}

Simplify

\frac{2 x ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )} = \frac{- \frac{3}{2}}{2 ( - \frac{1}{2} )}

Simplify

\frac{2 x ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )} : x

\frac{2 x ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )}

Remove parentheses:

(- a) = - a

= \frac{- 2 x \frac{1}{2}}{- 2 \cdot \frac{1}{2}}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{2 x \frac{1}{2}}{2 \cdot \frac{1}{2}}

Multiply

2 x \frac{1}{2} : x

2 x \frac{1}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2 x}{2}

Cancel the common factor:

2

= 1 \cdot x

Multiply:

1 \cdot x = x

= x

= \frac{x}{2 \cdot \frac{1}{2}}

Multiply

2 \cdot \frac{1}{2} : 1

2 \cdot \frac{1}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= \frac{x}{1}

Apply rule

\frac{a}{1} = a

= x

Simplify

\frac{- \frac{3}{2}}{2 ( - \frac{1}{2} )} : \frac{3}{2}

\frac{- \frac{3}{2}}{2 ( - \frac{1}{2} )}

Remove parentheses:

(- a) = - a

= \frac{- \frac{3}{2}}{- 2 \cdot \frac{1}{2}}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{\frac{3}{2}}{2 \cdot \frac{1}{2}}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{3}{2 \cdot 2 \cdot \frac{1}{2}}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3}{4 \cdot \frac{1}{2}}

Multiply

4 \cdot \frac{1}{2} : 2

4 \cdot \frac{1}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 4}{2}

Multiply the numbers:

1 \cdot 4 = 4

= \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2

= \frac{3}{2}

x = \frac{3}{2}

x = \frac{3}{2}

x = \frac{3}{2}

Verify solutions by plugging them into the original equations

Check the solutions by plugging them into

x^{2} - y^{2} = \frac{1}{2}

Remove the ones that don't agree with the equation.

Check the solution

x = \frac{3}{2}, y = - \frac{1}{2} :

True

x^{2} - y^{2} = \frac{1}{2}

Plug in

x = \frac{3}{2}, y = - \frac{1}{2}

(\frac{3}{2})^{2} - (- \frac{1}{2})^{2} = \frac{1}{2}

Refine

\frac{1}{2} = \frac{1}{2}

True

Check the solution

x = - \frac{3}{2}, y = \frac{1}{2} :

True

x^{2} - y^{2} = \frac{1}{2}

Plug in

x = - \frac{3}{2}, y = \frac{1}{2}

(- \frac{3}{2})^{2} - (\frac{1}{2})^{2} = \frac{1}{2}

Refine

\frac{1}{2} = \frac{1}{2}

True

Check the solutions by plugging them into

2 x y = - \frac{3}{2}

Remove the ones that don't agree with the equation.

Check the solution

x = \frac{3}{2}, y = - \frac{1}{2} :

True

2 x y = - \frac{3}{2}

Plug in

x = \frac{3}{2}, y = - \frac{1}{2}

2 \cdot \frac{3}{2} (- \frac{1}{2}) = - \frac{3}{2}

Refine

- \frac{3}{2} = - \frac{3}{2}

True

Check the solution

x = - \frac{3}{2}, y = \frac{1}{2} :

True

2 x y = - \frac{3}{2}

Plug in

x = - \frac{3}{2}, y = \frac{1}{2}

2 (- \frac{3}{2}) \frac{1}{2} = - \frac{3}{2}

Refine

- \frac{3}{2} = - \frac{3}{2}

True

Therefore, the final solutions for

x^{2} - y^{2} = \frac{1}{2}, 2 x y = - \frac{3}{2}

are

(x = - \frac{3}{2}, x = \frac{3}{2}, y = \frac{1}{2} y = - \frac{1}{2})

Substitute back

u = x + y i

u = - \frac{3}{2} + \frac{1}{2} i, u = \frac{3}{2} - \frac{1}{2} i

Solve

u^{2} = \frac{1}{2} + i \frac{3}{2} : u = \frac{3}{2} + \frac{1}{2} i, u = - \frac{3}{2} - \frac{1}{2} i

u^{2} = \frac{1}{2} + i \frac{3}{2}

Substitute

u = x + y i

(x + y i)^{2} = \frac{1}{2} + i \frac{3}{2}

Expand

(x + y i)^{2} : (x^{2} - y^{2}) + 2 i x y

(x + y i)^{2}

= (x + i y)^{2}

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = x, b = y i

= x^{2} + 2 x y i + (y i)^{2}

(y i)^{2} = - y^{2}

(y i)^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= i^{2} y^{2}

i^{2} = - 1

i^{2}

Apply imaginary number rule:

i^{2} = - 1

= - 1

= (- 1) y^{2}

Refine

= - y^{2}

= x^{2} + 2 i x y - y^{2}

Rewrite

x^{2} + 2 i x y - y^{2}

in standard complex form:

(x^{2} - y^{2}) + 2 x y i

x^{2} + 2 i x y - y^{2}

Group the real part and the imaginary part of the complex number

= (x^{2} - y^{2}) + 2 x y i

= (x^{2} - y^{2}) + 2 x y i

(x^{2} - y^{2}) + 2 i x y = \frac{1}{2} + i \frac{3}{2}

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[x^{2} - y^{2} = \frac{1}{2} 2 x y = \frac{3}{2}]

[x^{2} - y^{2} = \frac{1}{2} 2 x y = \frac{3}{2}] : (x = \frac{3}{2}, x = - \frac{3}{2}, y = \frac{1}{2} y = - \frac{1}{2})

[x^{2} - y^{2} = \frac{1}{2} 2 x y = \frac{3}{2}]

Isolate

x

for

2 x y = \frac{3}{2} : x = \frac{3}{4 y}

2 x y = \frac{3}{2}

Divide both sides by

2 y

2 x y = \frac{3}{2}

Divide both sides by

2 y

\frac{2 x y}{2 y} = \frac{\frac{3}{2}}{2 y}

Simplify

\frac{2 x y}{2 y} = \frac{\frac{3}{2}}{2 y}

Simplify

\frac{2 x y}{2 y} : x

\frac{2 x y}{2 y}

Divide the numbers:

\frac{2}{2} = 1

= \frac{x y}{y}

Cancel the common factor:

y

= x

Simplify

\frac{\frac{3}{2}}{2 y} : \frac{3}{4 y}

\frac{\frac{3}{2}}{2 y}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{3}{2 \cdot 2 y}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3}{4 y}

x = \frac{3}{4 y}

x = \frac{3}{4 y}

x = \frac{3}{4 y}

Plug the solutions

x = \frac{3}{4 y}

into

x^{2} - y^{2} = \frac{1}{2}

For

x^{2} - y^{2} = \frac{1}{2}

, subsitute

x

with

\frac{3}{4 y} : y = \frac{1}{2}, y = - \frac{1}{2}

For

x^{2} - y^{2} = \frac{1}{2}

, subsitute

x

with

\frac{3}{4 y}

(\frac{3}{4 y})^{2} - y^{2} = \frac{1}{2}

Solve

(\frac{3}{4 y})^{2} - y^{2} = \frac{1}{2} : y = \frac{1}{2}, y = - \frac{1}{2}

(\frac{3}{4 y})^{2} - y^{2} = \frac{1}{2}

Multiply by LCM

(\frac{3}{4 y})^{2} - y^{2} = \frac{1}{2}

Simplify

(\frac{3}{4 y})^{2} : \frac{3}{16 y ^{2}}

(\frac{3}{4 y})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{( 3 ) ^{2}}{( 4 y ) ^{2}}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

(4 y)^{2} = 4^{2} y^{2}

= \frac{( 3 ) ^{2}}{4 ^{2} y ^{2}}

(3)^{2} : 3

Apply radical rule:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 3^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 3

= \frac{3}{4 ^{2} y ^{2}}

4^{2} = 16

= \frac{3}{16 y ^{2}}

\frac{3}{16 y ^{2}} - y^{2} = \frac{1}{2}

Find Least Common Multiplier of

16 y^{2}, 2 : 16 y^{2}

16 y^{2}, 2

Lowest Common Multiplier (LCM)

Least Common Multiplier of

16, 2 : 16

16, 2

Least Common Multiplier (LCM)

Prime factorization of

16 : 2 \cdot 2 \cdot 2 \cdot 2

16

16

divides by

216 = 8 \cdot 2

= 2 \cdot 8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Multiply each factor the greatest number of times it occurs in either

16

2

= 2 \cdot 2 \cdot 2 \cdot 2

Multiply the numbers:

2 \cdot 2 \cdot 2 \cdot 2 = 16

= 16

Compute an expression comprised of factors that appear either in

16 y^{2}

2

= 16 y^{2}

Multiply by LCM=

16 y^{2}

\frac{3}{16 y ^{2}} \cdot 16 y^{2} - y^{2} \cdot 16 y^{2} = \frac{1}{2} \cdot 16 y^{2}

Simplify

\frac{3}{16 y ^{2}} \cdot 16 y^{2} - y^{2} \cdot 16 y^{2} = \frac{1}{2} \cdot 16 y^{2}

Simplify

\frac{3}{16 y ^{2}} \cdot 16 y^{2} : 3

\frac{3}{16 y ^{2}} \cdot 16 y^{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{3 \cdot 16 y ^{2}}{16 y ^{2}}

Cancel the common factor:

16

= \frac{3 y ^{2}}{y ^{2}}

Cancel the common factor:

y^{2}

= 3

Simplify

- y^{2} \cdot 16 y^{2} : - 16 y^{4}

- y^{2} \cdot 16 y^{2}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

y^{2} y^{2} = y^{2 + 2}

= - 16 y^{2 + 2}

Add the numbers:

2 + 2 = 4

= - 16 y^{4}

Simplify

\frac{1}{2} \cdot 16 y^{2} : 8 y^{2}

\frac{1}{2} \cdot 16 y^{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 16}{2} y^{2}

\frac{1 \cdot 16}{2} = 8

\frac{1 \cdot 16}{2}

Multiply the numbers:

1 \cdot 16 = 16

= \frac{16}{2}

Divide the numbers:

\frac{16}{2} = 8

= 8

= 8 y^{2}

3 - 16 y^{4} = 8 y^{2}

3 - 16 y^{4} = 8 y^{2}

3 - 16 y^{4} = 8 y^{2}

Solve

3 - 16 y^{4} = 8 y^{2} : y = \frac{1}{2}, y = - \frac{1}{2}

3 - 16 y^{4} = 8 y^{2}

Move

8 y^{2}

to the left side

3 - 16 y^{4} = 8 y^{2}

Subtract

8 y^{2}

from both sides

3 - 16 y^{4} - 8 y^{2} = 8 y^{2} - 8 y^{2}

Simplify

3 - 16 y^{4} - 8 y^{2} = 0

3 - 16 y^{4} - 8 y^{2} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

- 16 y^{4} - 8 y^{2} + 3 = 0

Rewrite the equation with

u = y^{2}

and

u^{2} = y^{4}

- 16 u^{2} - 8 u + 3 = 0

Solve

- 16 u^{2} - 8 u + 3 = 0 : u = - \frac{3}{4}, u = \frac{1}{4}

- 16 u^{2} - 8 u + 3 = 0

Solve with the quadratic formula

- 16 u^{2} - 8 u + 3 = 0

Quadratic Equation Formula:

For

a = - 16, b = - 8, c = 3

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 16 ) \cdot 3}{2 ( - 16 )}

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 16 ) \cdot 3}{2 ( - 16 )}

(- 8)^{2} - 4 (- 16) \cdot 3 = 16

(- 8)^{2} - 4 (- 16) \cdot 3

Apply rule

- (- a) = a

= (- 8)^{2} + 4 \cdot 16 \cdot 3

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 8)^{2} = 8^{2}

= 8^{2} + 4 \cdot 16 \cdot 3

Multiply the numbers:

4 \cdot 16 \cdot 3 = 192

= 8^{2} + 192

8^{2} = 64

= 64 + 192

Add the numbers:

64 + 192 = 256

= 256

Factor the number:

256 = 1 6^{2}

= 1 6^{2}

Apply radical rule:

1 6^{2} = 16

= 16

u_{1, 2} = \frac{- ( - 8 ) \pm 16}{2 ( - 16 )}

Separate the solutions

u_{1} = \frac{- ( - 8 ) + 16}{2 ( - 16 )}, u_{2} = \frac{- ( - 8 ) - 16}{2 ( - 16 )}

u = \frac{- ( - 8 ) + 16}{2 ( - 16 )} : - \frac{3}{4}

\frac{- ( - 8 ) + 16}{2 ( - 16 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{8 + 16}{- 2 \cdot 16}

Add the numbers:

8 + 16 = 24

= \frac{24}{- 2 \cdot 16}

Multiply the numbers:

2 \cdot 16 = 32

= \frac{24}{- 32}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{24}{32}

Cancel the common factor:

8

= - \frac{3}{4}

u = \frac{- ( - 8 ) - 16}{2 ( - 16 )} : \frac{1}{4}

\frac{- ( - 8 ) - 16}{2 ( - 16 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{8 - 16}{- 2 \cdot 16}

Subtract the numbers:

8 - 16 = - 8

= \frac{- 8}{- 2 \cdot 16}

Multiply the numbers:

2 \cdot 16 = 32

= \frac{- 8}{- 32}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{8}{32}

Cancel the common factor:

8

= \frac{1}{4}

The solutions to the quadratic equation are:

u = - \frac{3}{4}, u = \frac{1}{4}

u = - \frac{3}{4}, u = \frac{1}{4}

Substitute back

u = y^{2},

solve for

y

Solve

y^{2} = - \frac{3}{4} :

No Solution for

y \in R

y^{2} = - \frac{3}{4}

x^{2}

cannot be negative for

x \in R

No Solution for y \in R

Solve

y^{2} = \frac{1}{4} : y = \frac{1}{2}, y = - \frac{1}{2}

y^{2} = \frac{1}{4}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

y = \frac{1}{4}, y = - \frac{1}{4}

\frac{1}{4} = \frac{1}{2}

\frac{1}{4}

Apply radical rule:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= \frac{1}{4}

Apply radical rule:

1 = 1

1 = 1

= \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

a^{2} = a, a \geq 0

2^{2} = 2

= 2

= \frac{1}{2}

- \frac{1}{4} = - \frac{1}{2}

- \frac{1}{4}

Apply radical rule:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= - \frac{1}{4}

Apply radical rule:

1 = 1

1 = 1

= - \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

a^{2} = a, a \geq 0

2^{2} = 2

= 2

= - \frac{1}{2}

y = \frac{1}{2}, y = - \frac{1}{2}

The solutions are

y = \frac{1}{2}, y = - \frac{1}{2}

y = \frac{1}{2}, y = - \frac{1}{2}

Verify Solutions

Find undefined (singularity) points:

y = 0

Take the denominator(s) of

(\frac{3}{4 y})^{2} - y^{2}

and compare to zero

Solve

4 y = 0 : y = 0

4 y = 0

Divide both sides by

4

4 y = 0

Divide both sides by

4

\frac{4 y}{4} = \frac{0}{4}

Simplify

y = 0

y = 0

The following points are undefined

y = 0

Combine undefined points with solutions:

y = \frac{1}{2}, y = - \frac{1}{2}

Plug the solutions

y = \frac{1}{2}, y = - \frac{1}{2}

into

2 x y = \frac{3}{2}

For

2 x y = \frac{3}{2}

, subsitute

y

with

\frac{1}{2} : x = \frac{3}{2}

For

2 x y = \frac{3}{2}

, subsitute

y

with

\frac{1}{2}

2 x \frac{1}{2} = \frac{3}{2}

Solve

2 x \frac{1}{2} = \frac{3}{2} : x = \frac{3}{2}

2 x \frac{1}{2} = \frac{3}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

\frac{1 \cdot 2}{2} x = \frac{3}{2}

Cancel the common factor:

2

x \cdot 1 = \frac{3}{2}

Multiply:

x \cdot 1 = x

x = \frac{3}{2}

For

2 x y = \frac{3}{2}

, subsitute

y

with

- \frac{1}{2} : x = - \frac{3}{2}

For

2 x y = \frac{3}{2}

, subsitute

y

with

- \frac{1}{2}

2 x (- \frac{1}{2}) = \frac{3}{2}

Solve

2 x (- \frac{1}{2}) = \frac{3}{2} : x = - \frac{3}{2}

2 x (- \frac{1}{2}) = \frac{3}{2}

Divide both sides by

2 (- \frac{1}{2})

2 x (- \frac{1}{2}) = \frac{3}{2}

Divide both sides by

2 (- \frac{1}{2})

\frac{2 x ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )} = \frac{\frac{3}{2}}{2 ( - \frac{1}{2} )}

Simplify

\frac{2 x ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )} = \frac{\frac{3}{2}}{2 ( - \frac{1}{2} )}

Simplify

\frac{2 x ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )} : x

\frac{2 x ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )}

Remove parentheses:

(- a) = - a

= \frac{- 2 x \frac{1}{2}}{- 2 \cdot \frac{1}{2}}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{2 x \frac{1}{2}}{2 \cdot \frac{1}{2}}

Multiply

2 x \frac{1}{2} : x

2 x \frac{1}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2 x}{2}

Cancel the common factor:

2

= 1 \cdot x

Multiply:

1 \cdot x = x

= x

= \frac{x}{2 \cdot \frac{1}{2}}

Multiply

2 \cdot \frac{1}{2} : 1

2 \cdot \frac{1}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= \frac{x}{1}

Apply rule

\frac{a}{1} = a

= x

Simplify

\frac{\frac{3}{2}}{2 ( - \frac{1}{2} )} : - \frac{3}{2}

\frac{\frac{3}{2}}{2 ( - \frac{1}{2} )}

Remove parentheses:

(- a) = - a

= \frac{\frac{3}{2}}{- 2 \cdot \frac{1}{2}}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{\frac{3}{2}}{2 \cdot \frac{1}{2}}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

\frac{\frac{3}{2}}{2 \cdot \frac{1}{2}} = \frac{3}{2 \cdot 2 \cdot \frac{1}{2}}

= - \frac{3}{2 \cdot 2 \cdot \frac{1}{2}}

Multiply the numbers:

2 \cdot 2 = 4

= - \frac{3}{4 \cdot \frac{1}{2}}

Multiply

4 \cdot \frac{1}{2} : 2

4 \cdot \frac{1}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 4}{2}

Multiply the numbers:

1 \cdot 4 = 4

= \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2

= - \frac{3}{2}

x = - \frac{3}{2}

x = - \frac{3}{2}

x = - \frac{3}{2}

Verify solutions by plugging them into the original equations

Check the solutions by plugging them into

x^{2} - y^{2} = \frac{1}{2}

Remove the ones that don't agree with the equation.

Check the solution

x = - \frac{3}{2}, y = - \frac{1}{2} :

True

x^{2} - y^{2} = \frac{1}{2}

Plug in

x = - \frac{3}{2}, y = - \frac{1}{2}

(- \frac{3}{2})^{2} - (- \frac{1}{2})^{2} = \frac{1}{2}

Refine

\frac{1}{2} = \frac{1}{2}

True

Check the solution

x = \frac{3}{2}, y = \frac{1}{2} :

True

x^{2} - y^{2} = \frac{1}{2}

Plug in

x = \frac{3}{2}, y = \frac{1}{2}

(\frac{3}{2})^{2} - (\frac{1}{2})^{2} = \frac{1}{2}

Refine

\frac{1}{2} = \frac{1}{2}

True

Check the solutions by plugging them into

2 x y = \frac{3}{2}

Remove the ones that don't agree with the equation.

Check the solution

x = - \frac{3}{2}, y = - \frac{1}{2} :

True

2 x y = \frac{3}{2}

Plug in

x = - \frac{3}{2}, y = - \frac{1}{2}

2 (- \frac{3}{2}) (- \frac{1}{2}) = \frac{3}{2}

Refine

\frac{3}{2} = \frac{3}{2}

True

Check the solution

x = \frac{3}{2}, y = \frac{1}{2} :

True

2 x y = \frac{3}{2}

Plug in

x = \frac{3}{2}, y = \frac{1}{2}

2 \cdot \frac{3}{2} \cdot \frac{1}{2} = \frac{3}{2}

Refine

\frac{3}{2} = \frac{3}{2}

True

Therefore, the final solutions for

x^{2} - y^{2} = \frac{1}{2}, 2 x y = \frac{3}{2}

are

(x = \frac{3}{2}, x = - \frac{3}{2}, y = \frac{1}{2} y = - \frac{1}{2})

Substitute back

u = x + y i

u = \frac{3}{2} + \frac{1}{2} i, u = - \frac{3}{2} - \frac{1}{2} i

The solutions are

u = - \frac{3}{2} + \frac{1}{2} i, u = \frac{3}{2} - \frac{1}{2} i, u = \frac{3}{2} + \frac{1}{2} i, u = - \frac{3}{2} - \frac{1}{2} i

u = - \frac{3}{2} + \frac{1}{2} i, u = \frac{3}{2} - \frac{1}{2} i, u = \frac{3}{2} + \frac{1}{2} i, u = - \frac{3}{2} - \frac{1}{2} i

Substitute back

u = sec (3 θ)

sec (3 θ) = - \frac{3}{2} + \frac{1}{2} i, sec (3 θ) = \frac{3}{2} - \frac{1}{2} i, sec (3 θ) = \frac{3}{2} + \frac{1}{2} i, sec (3 θ) = - \frac{3}{2} - \frac{1}{2} i

sec (3 θ) = - \frac{3}{2} + \frac{1}{2} i, sec (3 θ) = \frac{3}{2} - \frac{1}{2} i, sec (3 θ) = \frac{3}{2} + \frac{1}{2} i, sec (3 θ) = - \frac{3}{2} - \frac{1}{2} i

sec (3 θ) = - \frac{3}{2} + \frac{1}{2} i :

No Solution

sec (3 θ) = - \frac{3}{2} + \frac{1}{2} i

No Solution

sec (3 θ) = \frac{3}{2} - \frac{1}{2} i :

No Solution

sec (3 θ) = \frac{3}{2} - \frac{1}{2} i

No Solution

sec (3 θ) = \frac{3}{2} + \frac{1}{2} i :

No Solution

sec (3 θ) = \frac{3}{2} + \frac{1}{2} i

No Solution

sec (3 θ) = - \frac{3}{2} - \frac{1}{2} i :

No Solution

sec (3 θ) = - \frac{3}{2} - \frac{1}{2} i

No Solution

Combine all the solutions

No Solution for θ \in R

Popular Examples

sin(x)=0.04

sin (x) = 0.04

sin(x)=0.24

sin (x) = 0.24

sin(x)=0.95

sin (x) = 0.95

sin^2(x)=2cos^2(x/2)

sin^{2} (x) = 2 cos^{2} (\frac{x}{2})

-cos(a)-11=2cos(a)-9

- cos (a) - 11 = 2 cos (a) - 9

Frequently Asked Questions (FAQ)

What is the general solution for sec^2(3θ)cos^2(3θ)=sec^2(3θ)+cos^2(3θ) ?
The general solution for sec^2(3θ)cos^2(3θ)=sec^2(3θ)+cos^2(3θ) is No Solution for θ\in\mathbb{R}