English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n, x = 21 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

\frac{1}{2}

from both sides

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

Simplify

cos^{4} (x) - sin^{4} (x) - \frac{1}{2} : \frac{2 cos ^{4} ( x ) - 2 sin ^{4} ( x ) - 1}{2}

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

Convert element to fraction:

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

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

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

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

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

Apply exponent rule:

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

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

Add the numbers:

1 + 1 = 2

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

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

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

Apply Perfect Square Formula:

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

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

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

Simplify

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

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

Multiply the numbers:

2 \cdot 1 = 2

= 2 sin^{2} (x)

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

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

Apply exponent rule:

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

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

Multiply the numbers:

2 \cdot 2 = 4

= sin^{4} (x)

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

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

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

Expand

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

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

Distribute parentheses

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

Apply minus-plus rules

+ (- a) = - a

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

Simplify

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

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

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

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

Add/Subtract the numbers:

- 1 + 2 = 1

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

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

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

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

Solve by substitution

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

Let:

sin (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 = sin (x)

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

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

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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

General solutions for

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

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}

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = - \frac{1}{2} : x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

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

General solutions for

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

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}

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Popular Examples

sin(x)+cos(x)=tan(x)+1

sin (x) + cos (x) = tan (x) + 1

cos(x)= 1/4 ,0<= x<= 2pi

cos (x) = \frac{1}{4}, 0 \leq x \leq 2 π

tan^2(x)+1=a^2

tan^{2} (x) + 1 = a^{2}

2=2cos(t)

2 = 2 cos (t)

50sin(45)=600sin(θ)

50 sin (4 5^{\circ}) = 600 sin (θ)