Solve $\sin (x)+\sin (2x)+\sin (3x)=0$ for $x\in [0,2\pi ]$.

Solution

1. Group the terms: $(\sin (3x)+\sin (x))+\sin (2x)=0$.
2. Apply the sum-to-product formula: $2\sin (2x)\cos (x)+\sin (2x)=0$.
3. Factor out $\sin (2x)$: $\sin (2x)(2\cos (x)+1)=0$.
4. Solve $\sin (2x)=0$ giving $x=0,\pi /2,\pi ,3\pi /2,2\pi$ and $\cos (x)=-1/2$ giving $x=2\pi /3,4\pi /3$.
5. Hence the answer is (B).