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

The general solution for -sin(2x)=sin(4x) is x=pin,x=(pi+2pin)/2 ,x=(pi(3n+1))/3 ,x=(pi(3n+2))/3

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

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

x = πn, x = \frac{π + 2 πn}{2}, x = \frac{π ( 3 n + 1 )}{3}, x = \frac{π ( 3 n + 2 )}{3}

Solution steps

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

Subtract

sin (4 x)

from both sides

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

Let:

u = 2 x

- sin (u) - sin (2 u) = 0

Rewrite using trig identities

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

Factor

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

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

Solving each part separately

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

sin (u) = 0 : u = 2 πn, u = π + 2 πn

2 cos (u) + 1 = 0 : u = \frac{2 π}{3} + 2 πn, u = \frac{4 π}{3} + 2 πn

Combine all the solutions

u = 2 πn, u = π + 2 πn, u = \frac{2 π}{3} + 2 πn, u = \frac{4 π}{3} + 2 πn

Substitute back

u = 2 x

2 x = 2 πn : x = πn

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

2 x = \frac{2 π}{3} + 2 πn : x = \frac{π ( 3 n + 1 )}{3}

2 x = \frac{4 π}{3} + 2 πn : x = \frac{π ( 3 n + 2 )}{3}

x = πn, x = \frac{π + 2 πn}{2}, x = \frac{π ( 3 n + 1 )}{3}, x = \frac{π ( 3 n + 2 )}{3}

sin (x) = - 0.61

2 cos (3 x) + 2 = 0

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

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

sin (θ) - 2 sin (θ) cos (θ) = 0

What is the general solution for -sin(2x)=sin(4x) ?
The general solution for -sin(2x)=sin(4x) is x=pin,x=(pi+2pin)/2 ,x=(pi(3n+1))/3 ,x=(pi(3n+2))/3