Explanation

(d)
We have, $a_{1} + a_{2}\cos{2x + a_{3}\sin^{2}{x = 0}}$ , for all $x$
$\Rightarrow a_{1} + a_{2}\cos{2x + a_{3}\left( \ \frac{1 - \cos{2x}}{2} \right) = 0},$ for all $x$
$\Rightarrow \left( \ a_{1} + \frac{a_{3}}{2} \right) + \left( a_{2} - \frac{a_{3}}{2} \right)\cos{2x = 0,\ \forall\ x}$
$\Rightarrow a_{1} + \frac{a_{3}}{2} = 0\$ and $a_{2} - \frac{a_{3}}{2} = 0$
$\Rightarrow a_{1} = - \frac{k}{2},a_{2} = \frac{k}{2},a_{3} = k$, where, $k \in R$
Hence, the solutions are $\left( \ - \frac{k}{2},\frac{k}{2},\ k \right)$, where $k$ is any real number
Thus, the number of triplets is infinite