What is the general solution for -sin(2x)+sin(x)=0 ?

The general solution for -sin(2x)+sin(x)=0 is x= pi/3+(4pin)/3 ,x=pi+(4pin)/3 ,x=4pin,x=2pi+4pin

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

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

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

Solution steps

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

Rewrite using trig identities

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

Solving each part separately

cos (\frac{3 x}{2}) = 0 or sin (\frac{x}{2}) = 0

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

sin (\frac{x}{2}) = 0 : x = 4 πn, x = 2 π + 4 πn

Combine all the solutions

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

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

cos (x) = \frac{1}{10}

u, x = 5 sinh (u)

sin (\frac{π}{2} x) + 1 = 0

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

What is the general solution for -sin(2x)+sin(x)=0 ?
The general solution for -sin(2x)+sin(x)=0 is x= pi/3+(4pin)/3 ,x=pi+(4pin)/3 ,x=4pin,x=2pi+4pin