What is the general solution for cos^2(x)+cos^4(x)+cos^6(x)=0 ?

The general solution for cos^2(x)+cos^4(x)+cos^6(x)=0 is x= pi/2+2pin,x=(3pi)/2+2pin

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cos^2(x)+cos^4(x)+cos^6(x)=0

Solution

cos^{2} (x) + cos^{4} (x) + cos^{6} (x) = 0

Solution

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

cos^{2} (x) + cos^{4} (x) + cos^{6} (x) = 0

Solve by substitution

cos^{2} (x) + cos^{4} (x) + cos^{6} (x) = 0

Let:

cos (x) = u

u^{2} + u^{4} + u^{6} = 0

u^{2} + u^{4} + u^{6} = 0 : u = 0, u = \frac{1}{2} + \frac{3}{2} i, u = - \frac{1}{2} - \frac{3}{2} i, u = - \frac{1}{2} + \frac{3}{2} i, u = \frac{1}{2} - \frac{3}{2} i

u^{2} + u^{4} + u^{6} = 0

Write in the standard form

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

u^{6} + u^{4} + u^{2} = 0

Rewrite the equation with

a = u^{2}, a^{2} = u^{4}

and

a^{3} = u^{6}

a^{3} + a^{2} + a = 0

Solve

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

a^{3} + a^{2} + a = 0

Factor

a^{3} + a^{2} + a : a (a^{2} + a + 1)

a^{3} + a^{2} + a

Apply exponent rule:

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

a^{2} = aa

= a^{2} a + aa + a

Factor out common term

a

= a (a^{2} + a + 1)

a (a^{2} + a + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

a = 0 or a^{2} + a + 1 = 0

Solve

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

a^{2} + a + 1 = 0

Solve with the quadratic formula

a^{2} + a + 1 = 0

Quadratic Equation Formula:

For

a = 1, b = 1, c = 1

a_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

a_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

Simplify

1^{2} - 4 \cdot 1 \cdot 1 : 3 i

1^{2} - 4 \cdot 1 \cdot 1

Apply rule

1^{a} = 1

1^{2} = 1

= 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

a_{1, 2} = \frac{- 1 \pm 3 i}{2 \cdot 1}

Separate the solutions

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

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

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

Multiply the numbers:

2 \cdot 1 = 2

= \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}

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

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

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

Multiply the numbers:

2 \cdot 1 = 2

= \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}

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

The solutions to the quadratic equation are:

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

The solutions are

a = 0, a = - \frac{1}{2} + i \frac{3}{2}, a = - \frac{1}{2} - i \frac{3}{2}

a = 0, a = - \frac{1}{2} + i \frac{3}{2}, a = - \frac{1}{2} - i \frac{3}{2}

Substitute back

a = u^{2},

solve for

u

Solve

u^{2} = 0 : u = 0

u^{2} = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

u = 0

Solve

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

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

Substitute

u = a + bi

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

Expand

(a + bi)^{2} : (a^{2} - b^{2}) + 2 iab

(a + bi)^{2}

= (a + ib)^{2}

Apply Perfect Square Formula:

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

a = a, b = bi

= a^{2} + 2 abi + (bi)^{2}

(bi)^{2} = - b^{2}

(bi)^{2}

Apply exponent rule:

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

= i^{2} b^{2}

i^{2} = - 1

i^{2}

Apply imaginary number rule:

i^{2} = - 1

= - 1

= (- 1) b^{2}

Refine

= - b^{2}

= a^{2} + 2 iab - b^{2}

Rewrite

a^{2} + 2 iab - b^{2}

in standard complex form:

(a^{2} - b^{2}) + 2 abi

a^{2} + 2 iab - b^{2}

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

= (a^{2} - b^{2}) + 2 abi

= (a^{2} - b^{2}) + 2 abi

(a^{2} - b^{2}) + 2 iab = - \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:

[a^{2} - b^{2} = - \frac{1}{2} 2 ab = \frac{3}{2}]

[a^{2} - b^{2} = - \frac{1}{2} 2 ab = \frac{3}{2}] : (a = \frac{1}{2}, a = - \frac{1}{2}, b = \frac{3}{2} b = - \frac{3}{2})

[a^{2} - b^{2} = - \frac{1}{2} 2 ab = \frac{3}{2}]

Isolate

a

for

2 ab = \frac{3}{2} : a = \frac{3}{4 b}

2 ab = \frac{3}{2}

Divide both sides by

2 b

2 ab = \frac{3}{2}

Divide both sides by

2 b

\frac{2 ab}{2 b} = \frac{\frac{3}{2}}{2 b}

Simplify

\frac{2 ab}{2 b} = \frac{\frac{3}{2}}{2 b}

Simplify

\frac{2 ab}{2 b} : a

\frac{2 ab}{2 b}

Divide the numbers:

\frac{2}{2} = 1

= \frac{ab}{b}

Cancel the common factor:

b

= a

Simplify

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

\frac{\frac{3}{2}}{2 b}

Apply the fraction rule:

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

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3}{4 b}

a = \frac{3}{4 b}

a = \frac{3}{4 b}

a = \frac{3}{4 b}

Plug the solutions

a = \frac{3}{4 b}

into

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

For

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

, subsitute

a

with

\frac{3}{4 b} : b = \frac{3}{2}, b = - \frac{3}{2}

For

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

, subsitute

a

with

\frac{3}{4 b}

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

Solve

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

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

Multiply by LCM

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

Simplify

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

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

Apply exponent rule:

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

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

Apply exponent rule:

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

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

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

4^{2} = 16

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

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

Find Least Common Multiplier of

16 b^{2}, 2 : 16 b^{2}

16 b^{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 b^{2}

2

= 16 b^{2}

Multiply by LCM=

16 b^{2}

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

Simplify

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

Simplify

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

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

Multiply fractions:

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

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

Cancel the common factor:

16

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

Cancel the common factor:

b^{2}

= 3

Simplify

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

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

Apply exponent rule:

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

b^{2} b^{2} = b^{2 + 2}

= - 16 b^{2 + 2}

Add the numbers:

2 + 2 = 4

= - 16 b^{4}

Simplify

- \frac{1}{2} \cdot 16 b^{2} : - 8 b^{2}

- \frac{1}{2} \cdot 16 b^{2}

Multiply fractions:

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

= - \frac{1 \cdot 16}{2} b^{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 b^{2}

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

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

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

Solve

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

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

Move

8 b^{2}

to the left side

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

Add

8 b^{2}

to both sides

3 - 16 b^{4} + 8 b^{2} = - 8 b^{2} + 8 b^{2}

Simplify

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

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

Write in the standard form

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

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

Rewrite the equation with

u = b^{2}

and

u^{2} = b^{4}

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

Solve

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

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{1}{4}

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

Remove parentheses:

(- a) = - a

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

Add/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}

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

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

Remove parentheses:

(- a) = - a

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

Subtract 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}

The solutions to the quadratic equation are:

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

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

Substitute back

u = b^{2},

solve for

b

Solve

b^{2} = - \frac{1}{4} :

No Solution for

b \in R

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

x^{2}

cannot be negative for

x \in R

No Solution for b \in R

Solve

b^{2} = \frac{3}{4} : b = \frac{3}{2}, b = - \frac{3}{2}

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

For

x^{2} = f (a)

the solutions are

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

b = \frac{3}{4}, b = - \frac{3}{4}

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

\frac{3}{4}

Apply radical rule:

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

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

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

- \frac{3}{4}

Apply radical rule:

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

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

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

The solutions are

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

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

Verify Solutions

Find undefined (singularity) points:

b = 0

Take the denominator(s) of

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

and compare to zero

Solve

4 b = 0 : b = 0

4 b = 0

Divide both sides by

4

4 b = 0

Divide both sides by

4

\frac{4 b}{4} = \frac{0}{4}

Simplify

b = 0

b = 0

The following points are undefined

b = 0

Combine undefined points with solutions:

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

Plug the solutions

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

into

2 ab = \frac{3}{2}

For

2 ab = \frac{3}{2}

, subsitute

b

with

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

For

2 ab = \frac{3}{2}

, subsitute

b

with

\frac{3}{2}

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

Solve

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

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

Multiply both sides by

2

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

Multiply both sides by

2

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

Simplify

23 a = 3

23 a = 3

Divide both sides by

23

23 a = 3

Divide both sides by

23

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

Simplify

a = \frac{1}{2}

a = \frac{1}{2}

For

2 ab = \frac{3}{2}

, subsitute

b

with

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

For

2 ab = \frac{3}{2}

, subsitute

b

with

- \frac{3}{2}

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

Solve

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

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

Divide both sides by

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

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

Divide both sides by

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

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

Simplify

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

Simplify

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

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

Simplify

\frac{2 a ( - \frac{3}{2} )}{2 ( - \frac{3}{2} )} : \frac{- 2 a \frac{3}{2}}{- 2 \cdot \frac{3}{2}}

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

Apply rule:

a (- b) = - ab

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

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

Apply rule:

a (- b) = - ab

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

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

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

Cancel the common factor:

- 2

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

Cancel the common factor:

\frac{3}{2}

= a

Simplify

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

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

Apply rule:

a (- b) = - ab

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

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

Apply the fraction rule:

\frac{a}{a} = 1

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

= \frac{1}{- 2}

Apply the fraction rule:

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

= - \frac{1}{2}

a = - \frac{1}{2}

a = - \frac{1}{2}

a = - \frac{1}{2}

Verify solutions by plugging them into the original equations

Check the solutions by plugging them into

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

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

Check the solution

a = - \frac{1}{2}, b = - \frac{3}{2} :

True

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

Plug in

a = - \frac{1}{2}, b = - \frac{3}{2}

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

Refine

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

True

Check the solution

a = \frac{1}{2}, b = \frac{3}{2} :

True

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

Plug in

a = \frac{1}{2}, b = \frac{3}{2}

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

Refine

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

True

Check the solutions by plugging them into

2 ab = \frac{3}{2}

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

Check the solution

a = - \frac{1}{2}, b = - \frac{3}{2} :

True

2 ab = \frac{3}{2}

Plug in

a = - \frac{1}{2}, b = - \frac{3}{2}

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

Refine

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

True

Check the solution

a = \frac{1}{2}, b = \frac{3}{2} :

True

2 ab = \frac{3}{2}

Plug in

a = \frac{1}{2}, b = \frac{3}{2}

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

Refine

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

True

Therefore, the final solutions for

a^{2} - b^{2} = - \frac{1}{2}, 2 ab = \frac{3}{2}

are

(a = \frac{1}{2}, a = - \frac{1}{2}, b = \frac{3}{2} b = - \frac{3}{2})

Substitute back

u = a + bi

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

Solve

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

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

Substitute

u = a + bi

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

Expand

(a + bi)^{2} : (a^{2} - b^{2}) + 2 iab

(a + bi)^{2}

= (a + ib)^{2}

Apply Perfect Square Formula:

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

a = a, b = bi

= a^{2} + 2 abi + (bi)^{2}

(bi)^{2} = - b^{2}

(bi)^{2}

Apply exponent rule:

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

= i^{2} b^{2}

i^{2} = - 1

i^{2}

Apply imaginary number rule:

i^{2} = - 1

= - 1

= (- 1) b^{2}

Refine

= - b^{2}

= a^{2} + 2 iab - b^{2}

Rewrite

a^{2} + 2 iab - b^{2}

in standard complex form:

(a^{2} - b^{2}) + 2 abi

a^{2} + 2 iab - b^{2}

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

= (a^{2} - b^{2}) + 2 abi

= (a^{2} - b^{2}) + 2 abi

(a^{2} - b^{2}) + 2 iab = - \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:

[a^{2} - b^{2} = - \frac{1}{2} 2 ab = - \frac{3}{2}]

[a^{2} - b^{2} = - \frac{1}{2} 2 ab = - \frac{3}{2}] : (a = - \frac{1}{2}, a = \frac{1}{2}, b = \frac{3}{2} b = - \frac{3}{2})

[a^{2} - b^{2} = - \frac{1}{2} 2 ab = - \frac{3}{2}]

Isolate

a

for

2 ab = - \frac{3}{2} : a = - \frac{3}{4 b}

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

Divide both sides by

2 b

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

Divide both sides by

2 b

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

Simplify

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

Simplify

\frac{2 ab}{2 b} : a

\frac{2 ab}{2 b}

Divide the numbers:

\frac{2}{2} = 1

= \frac{ab}{b}

Cancel the common factor:

b

= a

Simplify

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

\frac{- \frac{3}{2}}{2 b}

Apply the fraction rule:

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

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

Apply the fraction rule:

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

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

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

Multiply the numbers:

2 \cdot 2 = 4

= - \frac{3}{4 b}

a = - \frac{3}{4 b}

a = - \frac{3}{4 b}

a = - \frac{3}{4 b}

Plug the solutions

a = - \frac{3}{4 b}

into

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

For

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

, subsitute

a

with

- \frac{3}{4 b} : b = \frac{3}{2}, b = - \frac{3}{2}

For

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

, subsitute

a

with

- \frac{3}{4 b}

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

Solve

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

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

Multiply by LCM

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

Simplify

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

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

Apply exponent rule:

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

n

is even

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

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

Apply exponent rule:

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

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

Apply exponent rule:

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

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

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

4^{2} = 16

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

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

Find Least Common Multiplier of

16 b^{2}, 2 : 16 b^{2}

16 b^{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 b^{2}

2

= 16 b^{2}

Multiply by LCM=

16 b^{2}

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

Simplify

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

Simplify

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

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

Multiply fractions:

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

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

Cancel the common factor:

16

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

Cancel the common factor:

b^{2}

= 3

Simplify

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

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

Apply exponent rule:

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

b^{2} b^{2} = b^{2 + 2}

= - 16 b^{2 + 2}

Add the numbers:

2 + 2 = 4

= - 16 b^{4}

Simplify

- \frac{1}{2} \cdot 16 b^{2} : - 8 b^{2}

- \frac{1}{2} \cdot 16 b^{2}

Multiply fractions:

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

= - \frac{1 \cdot 16}{2} b^{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 b^{2}

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

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

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

Solve

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

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

Move

8 b^{2}

to the left side

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

Add

8 b^{2}

to both sides

3 - 16 b^{4} + 8 b^{2} = - 8 b^{2} + 8 b^{2}

Simplify

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

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

Write in the standard form

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

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

Rewrite the equation with

u = b^{2}

and

u^{2} = b^{4}

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

Solve

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

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{1}{4}

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

Remove parentheses:

(- a) = - a

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

Add/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}

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

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

Remove parentheses:

(- a) = - a

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

Subtract 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}

The solutions to the quadratic equation are:

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

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

Substitute back

u = b^{2},

solve for

b

Solve

b^{2} = - \frac{1}{4} :

No Solution for

b \in R

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

x^{2}

cannot be negative for

x \in R

No Solution for b \in R

Solve

b^{2} = \frac{3}{4} : b = \frac{3}{2}, b = - \frac{3}{2}

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

For

x^{2} = f (a)

the solutions are

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

b = \frac{3}{4}, b = - \frac{3}{4}

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

\frac{3}{4}

Apply radical rule:

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

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

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

- \frac{3}{4}

Apply radical rule:

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

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

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

The solutions are

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

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

Verify Solutions

Find undefined (singularity) points:

b = 0

Take the denominator(s) of

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

and compare to zero

Solve

4 b = 0 : b = 0

4 b = 0

Divide both sides by

4

4 b = 0

Divide both sides by

4

\frac{4 b}{4} = \frac{0}{4}

Simplify

b = 0

b = 0

The following points are undefined

b = 0

Combine undefined points with solutions:

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

Plug the solutions

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

into

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

For

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

, subsitute

b

with

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

For

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

, subsitute

b

with

\frac{3}{2}

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

Solve

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

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

Multiply both sides by

2

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

Multiply both sides by

2

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

Simplify

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

Simplify

2 \cdot 2 a \frac{3}{2} : 23 a

2 \cdot 2 a \frac{3}{2}

2 \cdot 2 = 2^{2}

2 \cdot 2

Apply exponent rule:

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

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

= 2^{1 + 1}

Add the numbers:

1 + 1 = 2

= 2^{2}

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

Apply the fraction rule:

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

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

Cancel

\frac{2 ^{2} a 3}{2} : 2 a 3

\frac{2 ^{2} a 3}{2}

\frac{2 ^{2}}{2} = 2

\frac{2 ^{2}}{2}

Apply exponent rule:

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

= 2^{2 - 1}

Subtract the numbers:

2 - 1 = 1

= 2^{1}

Apply exponent rule:

a^{1} = a

= 2

= 2 a 3

= 2 a 3

= 23 a

Simplify

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

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

Apply rule:

a (- b) = - ab

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

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

Convert

2

to fraction

: \frac{2}{1}

2

Convert element to fraction:

2 = \frac{2}{1}

= \frac{2}{1}

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

Cross-cancel common factor:

2

= - \frac{3}{1}

Apply the fraction rule:

\frac{a}{1} = a

= - 3

23 a = - 3

23 a = - 3

23 a = - 3

Divide both sides by

23

23 a = - 3

Divide both sides by

23

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

Simplify

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

Simplify

\frac{2 3 a}{2 3} : a

\frac{2 3 a}{2 3}

Cancel the common factor:

2

= \frac{3 a}{3}

Cancel the common factor:

3

= a

Simplify

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

\frac{- 3}{2 3}

Cancel the common factor:

3

= \frac{- 1}{2}

Apply the fraction rule:

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

= - \frac{1}{2}

a = - \frac{1}{2}

a = - \frac{1}{2}

a = - \frac{1}{2}

For

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

, subsitute

b

with

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

For

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

, subsitute

b

with

- \frac{3}{2}

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

Solve

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

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

Divide both sides by

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

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

Divide both sides by

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

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

Simplify

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

Simplify

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

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

Simplify

\frac{2 a ( - \frac{3}{2} )}{2 ( - \frac{3}{2} )} : \frac{- 2 a \frac{3}{2}}{- 2 \cdot \frac{3}{2}}

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

Apply rule:

a (- b) = - ab

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

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

Apply rule:

a (- b) = - ab

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

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

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

Cancel the common factor:

- 2

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

Cancel the common factor:

\frac{3}{2}

= a

Simplify

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

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

Apply rule:

a (- b) = - ab

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

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

Apply the fraction rule:

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

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

Apply the fraction rule:

\frac{a}{a} = 1

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

= \frac{1}{2}

a = \frac{1}{2}

a = \frac{1}{2}

a = \frac{1}{2}

Verify solutions by plugging them into the original equations

Check the solutions by plugging them into

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

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

Check the solution

a = \frac{1}{2}, b = - \frac{3}{2} :

True

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

Plug in

a = \frac{1}{2}, b = - \frac{3}{2}

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

Refine

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

True

Check the solution

a = - \frac{1}{2}, b = \frac{3}{2} :

True

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

Plug in

a = - \frac{1}{2}, b = \frac{3}{2}

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

Refine

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

True

Check the solutions by plugging them into

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

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

Check the solution

a = \frac{1}{2}, b = - \frac{3}{2} :

True

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

Plug in

a = \frac{1}{2}, b = - \frac{3}{2}

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

Refine

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

True

Check the solution

a = - \frac{1}{2}, b = \frac{3}{2} :

True

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

Plug in

a = - \frac{1}{2}, b = \frac{3}{2}

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

Refine

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

True

Therefore, the final solutions for

a^{2} - b^{2} = - \frac{1}{2}, 2 ab = - \frac{3}{2}

are

(a = - \frac{1}{2}, a = \frac{1}{2}, b = \frac{3}{2} b = - \frac{3}{2})

Substitute back

u = a + bi

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

The solutions are

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

Substitute back

u = cos (x)

cos (x) = 0, cos (x) = \frac{1}{2} + \frac{3}{2} i, cos (x) = - \frac{1}{2} - \frac{3}{2} i, cos (x) = - \frac{1}{2} + \frac{3}{2} i, cos (x) = \frac{1}{2} - \frac{3}{2} i

cos (x) = 0, cos (x) = \frac{1}{2} + \frac{3}{2} i, cos (x) = - \frac{1}{2} - \frac{3}{2} i, cos (x) = - \frac{1}{2} + \frac{3}{2} i, cos (x) = \frac{1}{2} - \frac{3}{2} i

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

General solutions for

cos (x) = 0

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

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

No Solution

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

No Solution

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

No Solution

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

No Solution

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

No Solution

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

No Solution

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

No Solution

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

No Solution

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Popular Examples

sin(x)=(-1)/4

sin (x) = \frac{- 1}{4}

sin^4(x)-sin^2(x)=0

sin^{4} (x) - sin^{2} (x) = 0

(cos(t)-4)(2sin^2(t)-1)=0

(cos (t) - 4) (2 sin^{2} (t) - 1) = 0

sin(75)= x/9

sin (7 5^{\circ}) = \frac{x}{9}

solvefor i,2cos^3(x)+sin(x)+1=2sin^2(x)solve for

i, 2 cos^{3} (x) + sin (x) + 1 = 2 sin^{2} (x)

Frequently Asked Questions (FAQ)

What is the general solution for cos^2(x)+cos^4(x)+cos^6(x)=0 ?
The general solution for cos^2(x)+cos^4(x)+cos^6(x)=0 is x= pi/2+2pin,x=(3pi)/2+2pin