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Popular >

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

Solution

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

Solution

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

Degrees

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n, x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

1

from both sides

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

Apply exponent rule:

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

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

Add the numbers:

1 + 1 = 2

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

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

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

Apply Perfect Square Formula:

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

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

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

Simplify

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

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

Multiply the numbers:

2 \cdot 1 = 2

= 2 cos^{2} (x)

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

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

Apply exponent rule:

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

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

Multiply the numbers:

2 \cdot 2 = 4

= cos^{4} (x)

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

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

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

Expand

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

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

Distribute parentheses

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

Apply minus-plus rules

+ (- a) = - a

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

Simplify

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

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

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

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

Add/Subtract the numbers:

- 1 + 2 = 1

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

1 - 4 u^{2} = 0

1 - 4 u^{2} = 0 : u = \frac{1}{2}, u = - \frac{1}{2}

1 - 4 u^{2} = 0

Move

1

to the right side

1 - 4 u^{2} = 0

Subtract

1

from both sides

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

Simplify

- 4 u^{2} = - 1

- 4 u^{2} = - 1

Divide both sides by

- 4

- 4 u^{2} = - 1

Divide both sides by

- 4

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

Simplify

u^{2} = \frac{1}{4}

u^{2} = \frac{1}{4}

For

x^{2} = f (a)

the solutions are

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

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

\frac{1}{4} = \frac{1}{2}

\frac{1}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{1}{2}

Apply rule

1 = 1

= \frac{1}{2}

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

- \frac{1}{4}

Simplify

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

\frac{1}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{1}{2}

= - \frac{1}{2}

Apply rule

1 = 1

= - \frac{1}{2}

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

Substitute back

u = cos (x)

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

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

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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

General solutions for

cos (x) = \frac{1}{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}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

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

General solutions for

cos (x) = - \frac{1}{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{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

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

Combine all the solutions

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

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