What is the general solution for sin(2x)-sin(x)=0,0<= x<2pi ?

The general solution for sin(2x)-sin(x)=0,0<= x<2pi is x=0,x=pi,x= pi/3 ,x=(5pi)/3

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sin(2x)-sin(x)=0,0<= x<2pi

sin (2 x) - sin (x) = 0, 0 \leq x < 2 π

x = 0, x = π, x = \frac{π}{3}, x = \frac{5 π}{3}

Solution steps

sin (2 x) - sin (x) = 0, 0 \leq x < 2 π

Rewrite using trig identities

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

Factor

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

sin (x) (2 cos (x) - 1) = 0

Solving each part separately

sin (x) = 0 or 2 cos (x) - 1 = 0

sin (x) = 0, 0 \leq x < 2 π : x = 0, x = π

2 cos (x) - 1 = 0, 0 \leq x < 2 π : x = \frac{π}{3}, x = \frac{5 π}{3}

Combine all the solutions

x = 0, x = π, x = \frac{π}{3}, x = \frac{5 π}{3}

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

2 sin^{2} (θ) + 3 cos (θ) = 3

2 sin (θ) = 1

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

tan (x - π) + 2 sin (x - π) = 0

What is the general solution for sin(2x)-sin(x)=0,0<= x<2pi ?
The general solution for sin(2x)-sin(x)=0,0<= x<2pi is x=0,x=pi,x= pi/3 ,x=(5pi)/3