What is the general solution for sin^2(x)-cos(x)= 1/2 ?

The general solution for sin^2(x)-cos(x)= 1/2 is x=1.19606…+2pin,x=2pi-1.19606…+2pin

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sin^2(x)-cos(x)= 1/2

Solution

sin^{2} (x) - cos (x) = \frac{1}{2}

Solution

x = 1.19606 \dots + 2 πn, x = 2 π - 1.19606 \dots + 2 πn

Degrees

x = 68.52929 \dots^{\circ} + 36 0^{\circ} n, x = 291.47070 \dots^{\circ} + 36 0^{\circ} n

Solution steps

sin^{2} (x) - cos (x) = \frac{1}{2}

Subtract

\frac{1}{2}

from both sides

sin^{2} (x) - cos (x) - \frac{1}{2} = 0

Simplify

sin^{2} (x) - cos (x) - \frac{1}{2} : \frac{2 sin ^{2} ( x ) - 2 cos ( x ) - 1}{2}

sin^{2} (x) - cos (x) - \frac{1}{2}

Convert element to fraction:

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

= \frac{sin ^{2} ( x ) \cdot 2}{2} - \frac{cos ( x ) \cdot 2}{2} - \frac{1}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{sin ^{2} ( x ) \cdot 2 - cos ( x ) \cdot 2 - 1}{2}

\frac{2 sin ^{2} ( x ) - 2 cos ( x ) - 1}{2} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

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

Rewrite using trig identities

- 1 - 2 cos (x) + 2 sin^{2} (x)

Use the Pythagorean identity:

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

sin^{2} (x) = 1 - cos^{2} (x)

= - 1 - 2 cos (x) + 2 (1 - cos^{2} (x))

Simplify

- 1 - 2 cos (x) + 2 (1 - cos^{2} (x)) : - 2 cos^{2} (x) - 2 cos (x) + 1

- 1 - 2 cos (x) + 2 (1 - cos^{2} (x))

Expand

2 (1 - cos^{2} (x)) : 2 - 2 cos^{2} (x)

2 (1 - cos^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 1, c = cos^{2} (x)

= 2 \cdot 1 - 2 cos^{2} (x)

Multiply the numbers:

2 \cdot 1 = 2

= 2 - 2 cos^{2} (x)

= - 1 - 2 cos (x) + 2 - 2 cos^{2} (x)

Simplify

- 1 - 2 cos (x) + 2 - 2 cos^{2} (x) : - 2 cos^{2} (x) - 2 cos (x) + 1

- 1 - 2 cos (x) + 2 - 2 cos^{2} (x)

Group like terms

= - 2 cos (x) - 2 cos^{2} (x) - 1 + 2

Add/Subtract the numbers:

- 1 + 2 = 1

= - 2 cos^{2} (x) - 2 cos (x) + 1

= - 2 cos^{2} (x) - 2 cos (x) + 1

= - 2 cos^{2} (x) - 2 cos (x) + 1

1 - 2 cos (x) - 2 cos^{2} (x) = 0

Solve by substitution

1 - 2 cos (x) - 2 cos^{2} (x) = 0

Let:

cos (x) = u

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

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

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 2, b = - 2, c = 1

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

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

(- 2)^{2} - 4 (- 2) \cdot 1 = 23

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

Apply rule

- (- a) = a

= (- 2)^{2} + 4 \cdot 2 \cdot 1

Apply exponent rule:

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

n

is even

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

= 2^{2} + 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 2^{2} + 8

2^{2} = 4

= 4 + 8

Add the numbers:

4 + 8 = 12

= 12

Prime factorization of

12 : 2^{2} \cdot 3

12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3

= 2^{2} \cdot 3

= 2^{2} \cdot 3

Apply radical rule:

= 3 2^{2}

Apply radical rule:

2^{2} = 2

= 23

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

Separate the solutions

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

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

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

Remove parentheses:

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

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

Multiply the numbers:

2 \cdot 2 = 4

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

Apply the fraction rule:

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

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

Cancel

\frac{2 + 2 3}{4} : \frac{1 + 3}{2}

\frac{2 + 2 3}{4}

Factor

2 + 23 : 2 (1 + 3)

2 + 23

Rewrite as

= 2 \cdot 1 + 23

Factor out common term

2

= 2 (1 + 3)

= \frac{2 ( 1 + 3 )}{4}

Cancel the common factor:

2

= \frac{1 + 3}{2}

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

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

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

Remove parentheses:

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

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

Multiply the numbers:

2 \cdot 2 = 4

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

Apply the fraction rule:

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

2 - 23 = - (23 - 2)

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

Factor

23 - 2 : 2 (3 - 1)

23 - 2

Rewrite as

= 23 - 2 \cdot 1

Factor out common term

2

= 2 (3 - 1)

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

Cancel the common factor:

2

= \frac{3 - 1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

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

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

No Solution

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

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = \frac{3 - 1}{2} : x = arccos (\frac{3 - 1}{2}) + 2 πn, x = 2 π - arccos (\frac{3 - 1}{2}) + 2 πn

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

Apply trig inverse properties

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

General solutions for

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

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{3 - 1}{2}) + 2 πn, x = 2 π - arccos (\frac{3 - 1}{2}) + 2 πn

x = arccos (\frac{3 - 1}{2}) + 2 πn, x = 2 π - arccos (\frac{3 - 1}{2}) + 2 πn

Combine all the solutions

x = arccos (\frac{3 - 1}{2}) + 2 πn, x = 2 π - arccos (\frac{3 - 1}{2}) + 2 πn

Show solutions in decimal form

x = 1.19606 \dots + 2 πn, x = 2 π - 1.19606 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for sin^2(x)-cos(x)= 1/2 ?
The general solution for sin^2(x)-cos(x)= 1/2 is x=1.19606…+2pin,x=2pi-1.19606…+2pin