Explanation

(d)
We have, $\cos{3x + \cos{2x = \sin{\frac{3x}{2} + \sin\frac{x}{2}}}}$
$\Rightarrow \ 2\cos{\frac{5x}{2}\cos{\frac{x}{2} = 2\sin{x\cos\frac{x}{2}}}}$
Either $\cos{\frac{x}{2} = 0}$
$\Rightarrow \ \frac{x}{2} = (2n + 1)\frac{\pi}{2}$
$\Rightarrow \ x = (2n + 1)\pi$
or $\cos{\frac{5x}{2} = \sin x}$
$\Rightarrow \ \cos{\frac{5x}{2} = \cos\left( \frac{\pi}{2} - x \right)}$
$\Rightarrow \ \frac{5x}{2} = 2n\pi \pm \left( \frac{\pi}{2} - x \right)$
Taking the +ve sign $\frac{7x}{2} = 2n\pi + \frac{\pi}{2}$
$\Rightarrow \ x = \frac{4n\pi}{7} + \frac{\pi}{7}$
Taking –ve sign
$\frac{3x}{2} = 2n\pi - \frac{\pi}{2}\ \Rightarrow \ x = \frac{4n\pi}{3} - \frac{\pi}{3}$
For $0 \leq x \leq 2\pi$
$x = \frac{\pi}{7},\frac{5\pi}{7},\frac{9\pi}{7},\frac{13\pi}{7},\ \pi$
Thus, number of solutions $= \ 5$