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=(2pi)/3 ,x=(4pi)/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{2 π}{3}, x = \frac{4 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{2 π}{3}, x = \frac{4 π}{3}

Combine all the solutions

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

2 sin (6 x) sin (3 x) = 1 - cos (9 x)

cos (θ) sin (θ) = 172245

sec (b) cot (b) = sin (b)

\frac{37}{sin ( x )} = \frac{30}{sin ( 3 0 ^{\circ} )}

\frac{1}{cos ( θ )} = 1

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