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=0,x=pi,x=2pi,x=(2pi)/3 ,x=(4pi)/3

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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 = 0, x = π, x = 2 π, x = \frac{2 π}{3}, x = \frac{4 π}{3}

Solution steps

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

2 cos (x) + 1 = 0, 0 \leq x \leq 2 π : x = \frac{2 π}{3}, x = \frac{4 π}{3}

Combine all the solutions

x = 0, x = π, x = 2 π, x = \frac{2 π}{3}, x = \frac{4 π}{3}

8 sin (2 x) = 16 cos (x)

3 arccos (x) = π

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

2 cos (4 x) + 5 = 6

cos (2 θ) = - \frac{3}{2}, 0 \leq θ \leq 2 π

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=0,x=pi,x=2pi,x=(2pi)/3 ,x=(4pi)/3