What is the general solution for sin(x)-sin(2x)=sin^3(x) ?

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

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sin(x)-sin(2x)=sin^3(x)

sin (x) - sin (2 x) = sin^{3} (x)

x = 2 πn, x = π + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Solution steps

sin (x) - sin (2 x) = sin^{3} (x)

Subtract

sin^{3} (x)

from both sides

sin (x) - sin (2 x) - sin^{3} (x) = 0

Rewrite using trig identities

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

Factor

sin (x) - sin^{3} (x) - 2 cos (x) sin (x) : sin (x) (1 - sin^{2} (x) - 2 cos (x))

sin (x) (1 - sin^{2} (x) - 2 cos (x)) = 0

Solving each part separately

sin (x) = 0 or 1 - sin^{2} (x) - 2 cos (x) = 0

sin (x) = 0 : x = 2 πn, x = π + 2 πn

1 - sin^{2} (x) - 2 cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

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

sin (x) = \frac{4}{5}, \frac{π}{2} \leq 0 < π, sin (2 x)

2 cos^{2} (θ) - 3 cos (θ) + 1 = 0, \frac{3 π}{2} < θ < 2 π

- 2 cos^{2} (x) + cos (x) = 0

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

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