What is the general solution for cos^4(2x)-sin^4(2x)=1 ?

The general solution for cos^4(2x)-sin^4(2x)=1 is x=pin,x= pi/2+pin

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cos^4(2x)-sin^4(2x)=1

Solution

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

Solution

x = πn, x = \frac{π}{2} + πn

Degrees

x = 0^{\circ} + 18 0^{\circ} n, x = 9 0^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

1

from both sides

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

Apply exponent rule:

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

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

Add the numbers:

1 + 1 = 2

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

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

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

Apply Perfect Square Formula:

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

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

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

Simplify

1^{2} - 2 \cdot 1 \cdot sin^{2} (2 x) + (sin^{2} (2 x))^{2} : 1 - 2 sin^{2} (2 x) + sin^{4} (2 x)

1^{2} - 2 \cdot 1 \cdot sin^{2} (2 x) + (sin^{2} (2 x))^{2}

Apply rule

1^{a} = 1

1^{2} = 1

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

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

2 \cdot 1 \cdot sin^{2} (2 x)

Multiply the numbers:

2 \cdot 1 = 2

= 2 sin^{2} (2 x)

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

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

Apply exponent rule:

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

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

Multiply the numbers:

2 \cdot 2 = 4

= sin^{4} (2 x)

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

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

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

- 1 + 1 = 0

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

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

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

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

Divide both sides by

- 2

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

Divide both sides by

- 2

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

Divide both sides by

- 2

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

Simplify

sin^{2} (2 x) = 0

sin^{2} (2 x) = 0

Apply rule

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

sin (2 x) = 0

General solutions for

sin (2 x) = 0

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

2 x = 0 + 2 πn, 2 x = π + 2 πn

2 x = 0 + 2 πn, 2 x = π + 2 πn

Solve

2 x = 0 + 2 πn : x = πn

2 x = 0 + 2 πn

0 + 2 πn = 2 πn

2 x = 2 πn

Divide both sides by

2

2 x = 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{2 πn}{2}

Simplify

x = πn

x = πn

Solve

2 x = π + 2 πn : x = \frac{π}{2} + πn

2 x = π + 2 πn

Divide both sides by

2

2 x = π + 2 πn

Divide both sides by

2

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

Simplify

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

x = πn, x = \frac{π}{2} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for cos^4(2x)-sin^4(2x)=1 ?
The general solution for cos^4(2x)-sin^4(2x)=1 is x=pin,x= pi/2+pin