The number of all possible triplets (a_1, a_2 | vedclass

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(D) Given the equation $a_1 + a_2 \cos 2x + a_3 \sin^2 x = 0$ for all $x$. Using the identity $\sin^2 x = \frac{1 - \cos 2x}{2}$,we substitute this in

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infinite

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(D) Given the equation $a_1 + a_2 \cos 2x + a_3 \sin^2 x = 0$ for all $x$. Using the identity $\sin^2 x = \frac{1 - \cos 2x}{2}$,we substitute this into the equation: $a_1 + a_2 \cos 2x + a_3 \left( \frac{1 - \cos 2x}{2} \right) = 0$ Rearranging the terms based on the coefficients of $\cos 2x$ and the constant term: $\left( a_1 + \frac{a_3}{2} \right) + \left( a_2 - \frac{a_3}{2} \right) \cos 2x = 0$ Since this must hold for all $x$,the coefficients must be zero: $a_1 + \frac{a_3}{2} = 0 \implies a_1 = -\frac{a_3}{2}$ $a_2 - \frac{a_3}{2} = 0 \implies a_2 = \frac{a_3}{2}$ Here,$a_3$ can be any real number. Thus,for every value of $a_3$,we get a unique triplet $(a_1, a_2, a_3)$. Since there are infinitely many real values for $a_3$,there are infinitely many such triplets.

The number of all possible triplets $(a_1, a_2, a_3)$ such that $a_1 + a_2 \cos 2x + a_3 \sin^2 x = 0$ for all $x$ is

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