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

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

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solvefor x,sin(4x)=-2(sin(2x))

Solution

solve for

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

Solution

x = πn, x = \frac{π + 2 πn}{2}

Degrees

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

Solution steps

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

Subtract

- 2 sin (2 x)

from both sides

sin (4 x) + 2 sin (2 x) = 0

Let:

u = 2 x

sin (2 u) + 2 sin (u) = 0

Rewrite using trig identities

sin (2 u) + 2 sin (u)

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

= 2 sin (u) cos (u) + 2 sin (u)

2 sin (u) + 2 cos (u) sin (u) = 0

Factor

2 sin (u) + 2 cos (u) sin (u) : 2 sin (u) (cos (u) + 1)

2 sin (u) + 2 cos (u) sin (u)

Factor out common term

2 sin (u)

= 2 sin (u) (1 + cos (u))

2 sin (u) (cos (u) + 1) = 0

Solving each part separately

sin (u) = 0 or cos (u) + 1 = 0

sin (u) = 0 : u = 2 πn, u = π + 2 πn

sin (u) = 0

General solutions for

sin (u) = 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}

u = 0 + 2 πn, u = π + 2 πn

u = 0 + 2 πn, u = π + 2 πn

Solve

u = 0 + 2 πn : u = 2 πn

u = 0 + 2 πn

0 + 2 πn = 2 πn

u = 2 πn

u = 2 πn, u = π + 2 πn

cos (u) + 1 = 0 : u = π + 2 πn

cos (u) + 1 = 0

Move

1

to the right side

cos (u) + 1 = 0

Subtract

1

from both sides

cos (u) + 1 - 1 = 0 - 1

Simplify

cos (u) = - 1

cos (u) = - 1

General solutions for

cos (u) = - 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}

u = π + 2 πn

u = π + 2 πn

Combine all the solutions

u = 2 πn, u = π + 2 πn

Substitute back

u = 2 x

2 x = 2 πn : x = πn

2 x = 2 πn

Divide both sides by

2

2 x = 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{2 πn}{2}

Simplify

x = πn

x = πn

2 x = π + 2 πn : x = \frac{π + 2 πn}{2}

2 x = π + 2 πn

Divide both sides by

2

2 x = π + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{π}{2} + \frac{2 πn}{2} : \frac{π + 2 πn}{2}

\frac{π}{2} + \frac{2 πn}{2}

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{π + 2 πn}{2}

x = \frac{π + 2 πn}{2}

x = \frac{π + 2 πn}{2}

x = \frac{π + 2 πn}{2}

x = πn, x = \frac{π + 2 πn}{2}

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Frequently Asked Questions (FAQ)

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