What is the general solution for 4sin(4x)=-8sin(2x) ?

The general solution for 4sin(4x)=-8sin(2x) is x=pin,x=(pi+2pin)/2

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4sin(4x)=-8sin(2x)

4 sin (4 x) = - 8 sin (2 x)

x = πn, x = \frac{π + 2 πn}{2}

Solution steps

4 sin (4 x) = - 8 sin (2 x)

Subtract

- 8 sin (2 x)

from both sides

4 sin (4 x) + 8 sin (2 x) = 0

Let:

u = 2 x

4 sin (2 u) + 8 sin (u) = 0

Rewrite using trig identities

8 sin (u) + 8 cos (u) sin (u) = 0

Factor

8 sin (u) + 8 cos (u) sin (u) : 8 sin (u) (cos (u) + 1)

8 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

cos (u) + 1 = 0 : 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 = π + 2 πn : x = \frac{π + 2 πn}{2}

x = πn, x = \frac{π + 2 πn}{2}

3 sin^{2} (θ) - cos^{2} (θ) = 0

2 cot (θ) - 3 = 0

\frac{1}{2} = cos (θ)

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

sin (2 x) cos (x) - sin (x) = 0

What is the general solution for 4sin(4x)=-8sin(2x) ?
The general solution for 4sin(4x)=-8sin(2x) is x=pin,x=(pi+2pin)/2