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

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

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

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

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

Solution steps

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

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, 0 \leq x < 2 π : x = \frac{π}{3}, x = π, x = \frac{5 π}{3}

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

Combine all the solutions

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

sin (2 x) = \frac{- 2 6}{7}

arctan (4 x) + arctan (6 x) = 90

2 sin (θ) tan (θ) - tan (θ) = 1 - 2 sin (θ)

7 cos (2 θ) - 9 = - 9 sin (θ)

1 - sin (x) = 3 cos (x)

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