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

The general solution for sin^4(x)=1-cos^4(x) is x=2pin,x=pi+2pin,x= pi/2+2pin,x=(3pi)/2+2pin

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

Solution

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

Solution

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

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

1 - cos^{4} (x)

from both sides

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

Apply exponent rule:

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

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

Rewrite using trig identities

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

Simplify

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

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

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

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

Add the numbers:

1 + 1 = 2

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

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

- 1 + 1 = 0

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Write in the standard form

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

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

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

2 v^{2} - 2 v = 0

Solve

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

2 v^{2} - 2 v = 0

Solve with the quadratic formula

2 v^{2} - 2 v = 0

Quadratic Equation Formula:

For

a = 2, b = - 2, c = 0

v_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

v_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

(- 2)^{2} - 4 \cdot 2 \cdot 0 = 2

(- 2)^{2} - 4 \cdot 2 \cdot 0

Apply exponent rule:

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

n

is even

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

= 2^{2} - 4 \cdot 2 \cdot 0

Apply rule

0 \cdot a = 0

= 2^{2} - 0

2^{2} - 0 = 2^{2}

= 2^{2}

Apply radical rule:

assuming

a \geq 0

= 2

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

2 + 2 = 4

= \frac{4}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

v = \frac{- ( - 2 ) - 2}{2 \cdot 2} : 0

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

Apply rule

- (- a) = a

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

Subtract the numbers:

2 - 2 = 0

= \frac{0}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{0}{4}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

The solutions to the quadratic equation are:

v = 1, v = 0

v = 1, v = 0

Substitute back

v = u^{2},

solve for

u

Solve

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

u^{2} = 1

For

x^{2} = f (a)

the solutions are

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

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

Solve

u^{2} = 0 : u = 0

u^{2} = 0

Apply rule

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

u = 0

The solutions are

u = 1, u = - 1, u = 0

Substitute back

u = cos (x)

cos (x) = 1, cos (x) = - 1, cos (x) = 0

cos (x) = 1, cos (x) = - 1, cos (x) = 0

cos (x) = 1 : x = 2 πn

cos (x) = 1

General solutions for

cos (x) = 1

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 = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

General solutions for

cos (x) = - 1

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 = π + 2 πn

x = π + 2 πn

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

cos (x) = 0

General solutions for

cos (x) = 0

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}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

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

Combine all the solutions

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

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Frequently Asked Questions (FAQ)

What is the general solution for sin^4(x)=1-cos^4(x) ?
The general solution for sin^4(x)=1-cos^4(x) is x=2pin,x=pi+2pin,x= pi/2+2pin,x=(3pi)/2+2pin