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solvefor x,sin^4(x)=(cos^2(x))/2

Solution

solve for

x, sin^{4} (x) = \frac{cos ^{2} ( x )}{2}

Solution

x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Degrees

x = 13 5^{\circ} + 36 0^{\circ} n, x = 22 5^{\circ} + 36 0^{\circ} n, x = 4 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n

Solution steps

sin^{4} (x) = \frac{cos ^{2} ( x )}{2}

Subtract

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

from both sides

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

Simplify

sin^{4} (x) - \frac{cos ^{2} ( x )}{2} : \frac{2 sin ^{4} ( x ) - cos ^{2} ( x )}{2}

sin^{4} (x) - \frac{cos ^{2} ( x )}{2}

Convert element to fraction:

sin^{4} (x) = \frac{sin ^{4} ( x ) 2}{2}

= \frac{sin ^{4} ( x ) \cdot 2}{2} - \frac{cos ^{2} ( x )}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{sin ^{4} ( x ) \cdot 2 - cos ^{2} ( x )}{2}

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

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

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

Factor

2 sin^{4} (x) - cos^{2} (x) : (2 sin^{2} (x) + cos (x)) (2 sin^{2} (x) - cos (x))

2 sin^{4} (x) - cos^{2} (x)

Rewrite

2 sin^{4} (x) - cos^{2} (x)

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

2 sin^{4} (x) - cos^{2} (x)

Apply radical rule:

a = (a)^{2}

2 = (2)^{2}

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

Apply exponent rule:

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

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

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

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

(2)^{2} (sin^{2} (x))^{2} = (2 sin^{2} (x))^{2}

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

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

Apply Difference of Two Squares Formula:

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

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

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

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

Solving each part separately

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

2 sin^{2} (x) + cos (x) = 0 : x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

2 sin^{2} (x) + cos (x) = 0

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Expand

u + (1 - u^{2}) 2 : u + 2 - 2 u^{2}

u + (1 - u^{2}) 2

= u + 2 (1 - u^{2})

Expand

2 (1 - u^{2}) : 2 - 2 u^{2}

2 (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 1, c = u^{2}

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

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

Multiply:

1 \cdot 2 = 2

= 2 - 2 u^{2}

= u + 2 - 2 u^{2}

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

1^{2} - 4 (- 2) 2 = 3

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 42 (- 2)

Apply rule

- (- a) = a

= 1 + 422

422 = 8

422

Apply radical rule:

a a = a

22 = 2

= 4 \cdot 2

Multiply the numbers:

4 \cdot 2 = 8

= 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

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

Separate the solutions

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

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

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

Remove parentheses:

(- a) = - a

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

Add/Subtract the numbers:

- 1 + 3 = 2

= \frac{2}{- 2 2}

Apply the fraction rule:

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

= - \frac{2}{2 2}

Divide the numbers:

\frac{2}{2} = 1

= - \frac{1}{2}

Rationalize

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

- \frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2}

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

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

Remove parentheses:

(- a) = - a

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

Subtract the numbers:

- 1 - 3 = - 4

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

Apply the fraction rule:

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

= \frac{4}{2 2}

Divide the numbers:

\frac{4}{2} = 2

= \frac{2}{2}

Apply radical rule:

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

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

Apply exponent rule:

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

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

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

Subtract the numbers:

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

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

Apply radical rule:

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

= 2

The solutions to the quadratic equation are:

u = - \frac{2}{2}, u = 2

Substitute back

u = cos (x)

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

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

cos (x) = - \frac{2}{2} : x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

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

General solutions for

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

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{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

cos (x) = 2 :

No Solution

cos (x) = 2

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

2 sin^{2} (x) - cos (x) = 0 : x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Expand

- u + (1 - u^{2}) 2 : - u + 2 - 2 u^{2}

- u + (1 - u^{2}) 2

= - u + 2 (1 - u^{2})

Expand

2 (1 - u^{2}) : 2 - 2 u^{2}

2 (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 1, c = u^{2}

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

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

Multiply:

1 \cdot 2 = 2

= 2 - 2 u^{2}

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

(- 1)^{2} - 4 (- 2) 2 = 3

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

Apply rule

- (- a) = a

= (- 1)^{2} + 422

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

422 = 8

422

Apply radical rule:

a a = a

22 = 2

= 4 \cdot 2

Multiply the numbers:

4 \cdot 2 = 8

= 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

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

Separate the solutions

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

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

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

Remove parentheses:

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

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

Add the numbers:

1 + 3 = 4

= \frac{4}{- 2 2}

Apply the fraction rule:

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

= - \frac{4}{2 2}

Divide the numbers:

\frac{4}{2} = 2

= \frac{2}{2}

Apply radical rule:

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

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

Apply exponent rule:

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

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

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

Subtract the numbers:

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

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

Apply radical rule:

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

= - 2

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

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

Remove parentheses:

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

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

Subtract the numbers:

1 - 3 = - 2

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

Apply the fraction rule:

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

= \frac{2}{2 2}

Divide the numbers:

\frac{2}{2} = 1

= \frac{1}{2}

Rationalize

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

\frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2}{2}

= \frac{2}{2}

The solutions to the quadratic equation are:

u = - 2, u = \frac{2}{2}

Substitute back

u = cos (x)

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

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

cos (x) = - 2 :

No Solution

cos (x) = - 2

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = \frac{2}{2} : x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

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

General solutions for

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

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}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Combine all the solutions

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Combine all the solutions

x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

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