The number of all possible triplets $(a_{1}, a_{2}, a_{3})$ such that $a_{1} + a_{2} c o s (2 x) + a_{3} s i n^{2} (x) = 0$ for all $x$ is

Zero

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One

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Three

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Infinite

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Solution

The correct option is D
Infinite

$a_{^{1}} + a_{^{2}} \cos (2 x) + a^{3} \sin^{2} (x) = 0$

$a_{1} + a_{2} (1 - 2 \sin^{2} x) + a_{3} = 0 a_{1} + a_{2} + a_{3} - 2 a_{2} \sin^{2} x = 0 a_{1} + a_{2} + a_{3} = 2 a_{2} \sin^{2} x S i n^{2} x = \frac{(a_{1} + a_{2} + a_{3})}{2 a^{2}}$
For all values of $x$ , there exists an infinite number of points between the interval $0 \leq \sin^{2} x \leq 1 .$
Thus, the number of triplets is infinite.
Hence, option D is correct.

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