We are solving the equation in the range $0 \leq x < 2\pi$:

$$\cos(\sin x) = \tfrac{1}{2}.$$

Solution:

For this equation to hold, we require: $$\sin x = \cos^{-1}\!\left(\tfrac{1}{2}\right).$$

Now, $$\cos^{-1}\!\left(\tfrac{1}{2}\right) = \pm \tfrac{\pi}{3}, \; \tfrac{5\pi}{3}, \; \tfrac{7\pi}{3}, \; \tfrac{11\pi}{3}, \ldots$$

However, since $$-1 \leq \sin x \leq 1,$$ the only possible values of $\sin x$ are within $[-1,1]$. But all candidate values above (such as $\tfrac{\pi}{3}$, $\tfrac{5\pi}{3}$, etc.) are greater than 1 in magnitude.

Thus, there is no solution for $\cos(\sin x) = \tfrac{1}{2}$.

Answer: (A) no solutions