Rewrite the equation as $\cos(\sin x) = \cos\left(\frac{\pi}{2} - \cos x\right)$. This implies that $\frac{\pi}{2} - \cos x = \pm \sin x + 2k\pi$ for some $k \in \mathbb{Z}$, so $\cos x \pm \sin x = \frac{\pi}{2} - 2k\pi$. Since $|\cos x \pm \sin x| \le |\cos x| + |\sin x| \le 2$, we have $k=0$, and so $\cos x \pm \sin x = \frac{\pi}{2}$. Squaring both sides of the equation we get $\cos^2 x \pm 2\cos x\sin x + \sin^2 x = \frac{\pi^2}{4}$ or $\pm \sin 2x = \frac{\pi^2}{4} - 1$. Since $\frac{\pi^2}{4} - 1 > \frac{9}{4} - 1 > 1$, this equation has no real solutions, so the initial equation does not have any real solutions either.