In the range \(0 \le x < 2\pi\) the equation \[\cos(\sin x) = \frac{1}{2}\] has

no solutions;

one solution;

two solutions;

three solutions.

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).