Trigonometry-Please Help me out

\hspace{-16}$Here $\bf{a_{1}+a_{2}.\cos(2x)+a_{3}.\sin^2(x)=0}$\\\\\\ $\bf{a_{1}+a_{2}.\cos(2x)+\frac{a_{3}}{2}(2\sin^2(x))=0}$\\\\\\ $\bf{a_{1}+a_{2}.\cos(2x)+\frac{a_{3}}{2}.\left\{1-\cos(2x)\right\}=0}$\\\\\\ $\bf{\left(a_{1}+\frac{a_{3}}{2}\right)+\left(a_{2}-\frac{a_{3}}{2}\right).\cos(2x)=0=0+0.\cos(2x)\;\forall x\in \mathbb{R}}$\\\\\\ So $\bf{\left(a_{1}+\frac{a_{3}}{2}\right)=0}$ and $\bf{\left(a_{2}-\frac{a_{3}}{2}\right)=0}$\\\\\\ Now Let $\bf{\frac{a_{3}}{2}=\lambda\Leftrightarrow a_{3}=2\lambda}$ Where $\bf{\lambda\in \mathbb{R}}$\\\\\\ So $\bf{a_{1}=-\lambda}$ and $\bf{a_{2}=\lambda}$ and $\bf{a_{3}=2\lambda}$\\\\\\ So $\bf{\left(a_{1}\;,a_{2}\;,a_{3}\right)=\left(-\lambda\;,\lambda\;,2\lambda\right)}$ Where $\bf{\lambda\in \mathbb{R}}$

we must have a_1 + a_2 = 0 (put x = 0)

a_1 - a_2 + a_3 = 0 (put x = pi/2)

2a_1 + a_3 = 0 (put x = pi/4)

so a_1 = -c/2, a_2 = c/2, a_3 = c for all real c.

Plug them in the equation, to see that it works for all x.

So d) infinite would be the answer.

Nice Shubhodip.