The number of all possible triplets $\left(a_{1}, a_{2}, a_{3}\right)$ such that $a_{1}+a_{2} \cos (

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The number of all possible triplets $\left(a_{1}, a_{2}, a_{3}ight)$ such that $a_{1}+a_{2} \cos (2 x)+a_{3} \sin ^{2} x=0 \forall x$ is\

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(d): Since $a_{1}+a_{2} \cos (2 x)+a_{3} \sin ^{2} x=0 \forall x$ Putting $x=0$ and $\frac{\pi}{2}$, we get\ $a_{1}+a_{2}=0$ and $a_{1}-a_{2}+a_{3}=0$\ $\Rightarrow \quad a_{2}=-a_{1}, a_{3}=a_{2}-a_{1}=-2 a_{1}$\ and the given equation becomes\ $a_{1}-a_{1} \cos 2 x-2 a_{1} \sin ^{2} x=0 \forall x$\ $\Rightarrow \quad a_{1}\left(1-\cos 2 x-2 \sin ^{2} xight)=0 \forall x$, $$ \left(\because 1-\cos ^{2} x=2 \sin ^{2} xight) $$ which is satisfied by all values of $a_{1}$.\ Hence, infinite number of triplets ( $a_{1},-a_{1},-2 a_{1}$ ) are possible.\