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

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

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

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

x = πn, x = \frac{π + 2 πn}{2}

Solution steps

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

Subtract

- 4 sin (2 x)

from both sides

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

Let:

u = 2 x

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

Rewrite using trig identities

4 sin (u) + 4 cos (u) sin (u) = 0

Factor

4 sin (u) + 4 cos (u) sin (u) : 4 sin (u) (cos (u) + 1)

4 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}

cos (x) = (\frac{5}{3}), cot (2 x)

\frac{sec ( x )}{tan ( x )} = \frac{cos ( x )}{( sin ^{2} ( x ) )}

x, tan (x) = \frac{1}{8}

sin (x) + 7 = 7

cos (3 x) = 0, 0 < x < π

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