Find all solutions of $\sin 2x = \sin x$ for $0^\circ \le x < 360^\circ$.
A$0^\circ, 60^\circ, 180^\circ, 300^\circ$CORRECT
B$0^\circ, 30^\circ, 180^\circ, 330^\circ$
C$60^\circ, 300^\circ$
D$0^\circ, 180^\circ$
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Why the answer is A, and why the others tempt you.
Using $\sin 2x = 2\sin x\cos x$: $2\sin x\cos x - \sin x = 0$, so $\sin x(2\cos x - 1) = 0$. Either $\sin x = 0$, giving $x = 0^\circ$ or $180^\circ$, or $\cos x = \dfrac{1}{2}$, giving $x = 60^\circ$ or $300^\circ$. Dividing both sides by $\sin x$ at the start loses the solutions $0^\circ$ and $180^\circ$.
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