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

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

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Popular >

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

Solution

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

Move

2 u^{2}

to the left side

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

Add

2 u^{2}

to both sides

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

Simplify

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

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

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

v^{2} + 2 v = 0

Solve

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

v^{2} + 2 v = 0

Solve with the quadratic formula

v^{2} + 2 v = 0

Quadratic Equation Formula:

For

a = 1, b = 2, c = 0

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

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

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

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

Separate the solutions

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

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

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

Add/Subtract the numbers:

- 2 + 2 = 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{- 2 - 2}{2 \cdot 1} : - 2

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

Subtract the numbers:

- 2 - 2 = - 4

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 4}{2}

Apply the fraction rule:

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

= - \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= - 2

The solutions to the quadratic equation are:

v = 0, v = - 2

v = 0, v = - 2

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} = - 2 : u = 2 i, u = - 2 i

u^{2} = - 2

For

x^{2} = f (a)

the solutions are

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

u = - 2, u = - - 2

Simplify

- 2 : 2 i

- 2

Apply radical rule:

- a = - 1 a

- 2 = - 1 2

= - 1 2

Apply imaginary number rule:

- 1 = i

= 2 i

Simplify

- - 2 : - 2 i

- - 2

Simplify

- 2 : 2 i

- 2

Apply radical rule:

- a = - 1 a

- 2 = - 1 2

= - 1 2

Apply imaginary number rule:

- 1 = i

= 2 i

= - 2 i

u = 2 i, u = - 2 i

The solutions are

u = 0, u = 2 i, u = - 2 i

Substitute back

u = sin (x)

sin (x) = 0, sin (x) = 2 i, sin (x) = - 2 i

sin (x) = 0, sin (x) = 2 i, sin (x) = - 2 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) = 2 i :

No Solution

sin (x) = 2 i

No Solution

sin (x) = - 2 i :

No Solution

sin (x) = - 2 i

No Solution

Combine all the solutions

x = 2 πn, x = π + 2 πn

Popular Examples

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3 tan (4 x) = 2

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- cos (2 x) = 0

tan(2x)=-(sqrt(3))/3

tan (2 x) = - \frac{3}{3}

2cot^2(x)-3csc(x)=0

2 cot^{2} (x) - 3 csc (x) = 0

3-4cos^2(x)=0

3 - 4 cos^{2} (x) = 0

Frequently Asked Questions (FAQ)

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