How many solutions does $\cos (\sin x)=\frac{1}{2}$ have?

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The equation \(\cos y = \dfrac{1}{2}\) has solutions \[y = \pm \frac{\pi}{3}, \pm \frac{5\pi}{3}, \pm \frac{7\pi}{3}, \pm \frac{11\pi}{3}, \dotsc .\]

So if \(\cos(\sin x) = \dfrac{1}{2}\) then \[\sin x = \pm \frac{\pi}{3}, \pm \frac{5\pi}{3}, \pm \frac{7\pi}{3}, \pm \frac{11\pi}{3}, \dotsc .\]

Now as \(\pi > 3\), we have \(\dfrac{\pi}{3} > 1\).

Therefore, all numbers on this list have magnitude greater than \(1\), and so lie outside the range of sine (\(-1 \le \sin x \le 1\)).

This means that \(\sin x\) cannot equal any of the numbers on this list, so there are no solutions.

The answer is (a).