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Popular >

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

Solution

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

Solution

x = 0.87161 \dots + 2 πn, x = 2 π - 0.87161 \dots + 2 πn, x = 2.26998 \dots + 2 πn, x = - 2.26998 \dots + 2 πn

Degrees

x = 49.93964 \dots^{\circ} + 36 0^{\circ} n, x = 310.06035 \dots^{\circ} + 36 0^{\circ} n, x = 130.06035 \dots^{\circ} + 36 0^{\circ} n, x = - 130.06035 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

cos^{4} (x)

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

= - cos^{2} (x) - cos^{4} (x) + 1 - cos^{2} (x)

Simplify

- cos^{2} (x) - cos^{4} (x) + 1 - cos^{2} (x) : - 2 cos^{2} (x) - cos^{4} (x) + 1

- cos^{2} (x) - cos^{4} (x) + 1 - cos^{2} (x)

Group like terms

= - cos^{2} (x) - cos^{4} (x) - cos^{2} (x) + 1

Add similar elements:

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

= - 2 cos^{2} (x) - cos^{4} (x) + 1

= - 2 cos^{2} (x) - cos^{4} (x) + 1

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Write in the standard form

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

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

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

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

Solve

- v^{2} - 2 v + 1 = 0 : v = - 1 - 2, v = 2 - 1

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

(- 2)^{2} - 4 (- 1) \cdot 1 = 22

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 2^{2} + 4

2^{2} = 4

= 4 + 4

Add the numbers:

4 + 4 = 8

= 8

Prime factorization of

8 : 2^{3}

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

Apply exponent rule:

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

= 2^{2} \cdot 2

Apply radical rule:

= 2 2^{2}

Apply radical rule:

2^{2} = 2

= 22

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

Separate the solutions

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

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

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

Remove parentheses:

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

Apply the fraction rule:

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

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

Cancel

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

\frac{2 + 2 2}{2}

Factor

2 + 22 : 2 (1 + 2)

2 + 22

Rewrite as

= 2 \cdot 1 + 22

Factor out common term

2

= 2 (1 + 2)

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

Divide the numbers:

\frac{2}{2} = 1

= 1 + 2

= - (1 + 2)

Distribute parentheses

= - (1) - (2)

Apply minus-plus rules

+ (- a) = - a

= - 1 - 2

v = \frac{- ( - 2 ) - 2 2}{2 ( - 1 )} : 2 - 1

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

Remove parentheses:

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

Apply the fraction rule:

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

2 - 22 = - (22 - 2)

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

Factor

22 - 2 : 2 (2 - 1)

22 - 2

Rewrite as

= 22 - 2 \cdot 1

Factor out common term

2

= 2 (2 - 1)

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

Divide the numbers:

\frac{2}{2} = 1

= 2 - 1

The solutions to the quadratic equation are:

v = - 1 - 2, v = 2 - 1

v = - 1 - 2, v = 2 - 1

Substitute back

v = u^{2},

solve for

u

Solve

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

u^{2} = - 1 - 2

For

x^{2} = f (a)

the solutions are

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

u = - 1 - 2, u = - - 1 - 2

Simplify

- 1 - 2 : i 1 + 2

- 1 - 2

Apply imaginary number rule:

- a = i a

= i 1 + 2

Simplify

- - 1 - 2 : - i 1 + 2

- - 1 - 2

Apply imaginary number rule:

- a = i a

= - i 1 + 2

u = i 1 + 2, u = - i 1 + 2

Solve

u^{2} = 2 - 1 : u = 2 - 1, u = - 2 - 1

u^{2} = 2 - 1

For

x^{2} = f (a)

the solutions are

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

u = 2 - 1, u = - 2 - 1

The solutions are

u = i 1 + 2, u = - i 1 + 2, u = 2 - 1, u = - 2 - 1

Substitute back

u = cos (x)

cos (x) = i 1 + 2, cos (x) = - i 1 + 2, cos (x) = 2 - 1, cos (x) = - 2 - 1

cos (x) = i 1 + 2, cos (x) = - i 1 + 2, cos (x) = 2 - 1, cos (x) = - 2 - 1

cos (x) = i 1 + 2 :

No Solution

cos (x) = i 1 + 2

No Solution

cos (x) = - i 1 + 2 :

No Solution

cos (x) = - i 1 + 2

No Solution

cos (x) = 2 - 1 : x = arccos (2 - 1) + 2 πn, x = 2 π - arccos (2 - 1) + 2 πn

cos (x) = 2 - 1

Apply trig inverse properties

cos (x) = 2 - 1

General solutions for

cos (x) = 2 - 1

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

x = arccos (2 - 1) + 2 πn, x = 2 π - arccos (2 - 1) + 2 πn

x = arccos (2 - 1) + 2 πn, x = 2 π - arccos (2 - 1) + 2 πn

cos (x) = - 2 - 1 : x = arccos (- 2 - 1) + 2 πn, x = - arccos (- 2 - 1) + 2 πn

cos (x) = - 2 - 1

Apply trig inverse properties

cos (x) = - 2 - 1

General solutions for

cos (x) = - 2 - 1

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

x = arccos (- 2 - 1) + 2 πn, x = - arccos (- 2 - 1) + 2 πn

x = arccos (- 2 - 1) + 2 πn, x = - arccos (- 2 - 1) + 2 πn

Combine all the solutions

x = arccos (2 - 1) + 2 πn, x = 2 π - arccos (2 - 1) + 2 πn, x = arccos (- 2 - 1) + 2 πn, x = - arccos (- 2 - 1) + 2 πn

Show solutions in decimal form

x = 0.87161 \dots + 2 πn, x = 2 π - 0.87161 \dots + 2 πn, x = 2.26998 \dots + 2 πn, x = - 2.26998 \dots + 2 πn

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