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,x=2pi

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

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

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

Solution steps

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

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

Combine all the solutions

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

tan (θ) = \frac{1}{4}

tan (θ) = \frac{1}{3}

2 sec (x) + 2 = 0

6 sin (θ) - 3 = 0

cot^{2} (x) - csc (x) = 1

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,x=2pi