Explanation

(b)
We know that,
$- \sqrt{a^{2} + b^{2}} \leq a\cos{\theta + b\sin\theta} \leq \sqrt{a^{2} + b^{2}}$
$\therefore\ - \sqrt{3} + 1 \leq \sqrt{3}\sin{x + \cos x} \leq \sqrt{3 + 1}$
$\Rightarrow \ - 2 \leq \sqrt{3}\sin{x + \cos x} \leq 2$
But, $\sqrt{3}\sin{x + \cos{x = 4}}$
Hence, given equation has no solution