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Question

If $0\le x<\dfrac\pi2$, then the number of values of $x$ for which $\sin x-\sin2x+\sin3x=0$ is:

Accepted Answer

$$\sin x - 2\sin x\cos x + 3\sin x - 4\sin^3x = 0 \implies 4\sin x - 4\sin^3x - 2\sin x\cos x = 0$$

$$2\sin x(1-\sin^2x) - \sin x\cos x = 0 \implies \sin x\cos x(2\cos x-1)=0$$

So $\sin x=0$, $\cos x=0$, or $\cos x=\tfrac12$. Within $\left[0,\tfrac\pi2\right)$: $x=0$ and $x=\tfrac\pi3$ --- 2 solutions.

Solutions · 1