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

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

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Popular >

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

Solution

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

Solution

x = 2 πn, x = π + 2 πn

Degrees

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

Solution steps

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

Switch sides

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

Move

u^{2}

to the left side

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

Add

u^{2}

to both sides

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

Simplify

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

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

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

v^{2} + v = 0

Solve

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

v^{2} + v = 0

Solve with the quadratic formula

v^{2} + v = 0

Quadratic Equation Formula:

For

a = 1, b = 1, c = 0

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

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

1^{2} - 4 \cdot 1 \cdot 0 = 1

1^{2} - 4 \cdot 1 \cdot 0

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 1 - 0

Subtract the numbers:

1 - 0 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

v = \frac{- 1 + 1}{2 \cdot 1} : 0

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

Add/Subtract the numbers:

- 1 + 1 = 0

= \frac{0}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{2}

Apply rule

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

= 0

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

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

Subtract the numbers:

- 1 - 1 = - 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{2}

Apply the fraction rule:

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

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

v = 0, v = - 1

v = 0, v = - 1

Substitute back

v = u^{2},

solve for

u

Solve

u^{2} = 0 : u = 0

u^{2} = 0

Apply rule

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

u = 0

Solve

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

u^{2} = - 1

For

x^{2} = f (a)

the solutions are

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

u = - 1, u = - - 1

Simplify

- 1 : i

- 1

Apply imaginary number rule:

- 1 = i

= i

Simplify

- - 1 : - i

- - 1

Apply imaginary number rule:

- 1 = i

= - i

u = i, u = - i

The solutions are

u = 0, u = i, u = - i

Substitute back

u = sin (x)

sin (x) = 0, sin (x) = i, sin (x) = - i

sin (x) = 0, sin (x) = i, sin (x) = - i

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (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}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 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, x = π + 2 πn

sin (x) = i :

No Solution

sin (x) = i

No Solution

sin (x) = - i :

No Solution

sin (x) = - i

No Solution

Combine all the solutions

x = 2 πn, x = π + 2 πn

Popular Examples

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

sin^2(θ)-cos^2(θ)=cos(θ)

sin^{2} (θ) - cos^{2} (θ) = cos (θ)

csc(t)=2

csc (t) = 2

arccosh(θ)=0.86806

arccosh (θ) = 0.86806

sin(t)= 5/13

sin (t) = \frac{5}{13}

Frequently Asked Questions (FAQ)

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