(d)

The given equation

$\sin \left(\pi \sin ^2 \theta\right)+\sin \left(\pi \cos ^2 \theta\right)=2 \cos \left(\frac{\pi}{2} \cos \theta\right)$

$\Rightarrow 2 \sin \left(\frac{\pi\left(\sin ^2 \theta+\cos ^2 \theta\right)}{2}\right)$ $\cos \left(\frac{\pi\left(\sin ^2 \theta-\cos ^2 \theta\right)}{2}\right)=2 \cos \left(\frac{\pi}{2} \cos \theta\right)$

$\Rightarrow \cos \left(\frac{\pi}{2} \cos 2 \theta\right)=\cos \left(\frac{\pi}{2} \cos \theta\right)$

$\Rightarrow \frac{\pi}{2} \cos 2 \theta=2 n \pi \pm \frac{\pi}{2} \cos \theta, n \in \text { Integers }$

$\Rightarrow \cos 2 \theta=4 n \pm \cos \theta, n \in I$ \\

$\text { Case I }$

$\text { If } \cos 2 \theta=4 n+\cos \theta \Rightarrow \cos 2 \theta-\cos \theta=4 n$

$\text { The above equation will hold only if } n=0$

$\text { so }$

$\Rightarrow 2 \cos 2 \theta-\cos \theta-1=0 \Rightarrow \cos \theta=1,-\frac{1}{2}$

$\Rightarrow \quad \theta=2 k \pi, 2 k \pi \pm \frac{2 \pi}{3}, k \in I$

$\because \quad \theta \in[0,2 \pi]$

So, $\quad \theta=0,2 \pi, \frac{2 \pi}{3}, \frac{4 \pi}{3}$

Case $II$

$\text { If } \cos 2 \theta=4 n-\cos \theta$

$\Rightarrow \quad \cos 2 \theta+\cos \theta=4 n$

The above equation will hold only if $n=0$,

$\text { so } \quad \cos 2 \theta+\cos \theta=0$

$\Rightarrow 2 \cos ^2 \theta+\cos \theta-1=0$

$\Rightarrow \quad \cos \theta=-1, \frac{1}{2}$

$\Rightarrow \quad \theta=(2 m+1) \pi, 2 m \pi \pm \frac{\pi}{3}, m \in I$

$\therefore$ There are total $7$ solutions.