The number of all the possible triplets $a_{1}, a_{2}, a_{3}$ such that $a_{1} + a_{2} cos (2 x) + a_{3} {sin}^{2} (x) = 0$ for all $x \in R$ is

0

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1

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3

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infinite

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Solution

The correct option is D infinite
$a_{1} + a_{2} cos (2 x) + a_{3} {sin}^{2} (x) = 0 \Rightarrow a_{1} + a_{2} cos (2 x) + a_{3} (\frac{1 - cos 2 x}{2}) = 0 \Rightarrow (a_{1} + \frac{a_{3}}{2}) + (a_{2} - \frac{a_{3}}{2}) cos 2 x = 0$

The above equation is valid for all $x \in R$ , so
$a_{1} + \frac{a_{3}}{2} = 0 and a_{2} - \frac{a_{3}}{2} = 0 \Rightarrow a_{3} = - 2 a_{1}, a_{2} = - a_{1}$

Hence, infinite number of triplets of $(a_{1}, - a_{1}, - 2 a_{1})$ is possible.

Alternate Solution:
Since the given equation is possible for all $x \in R$ , so
Putting $x = 0$ , we get
$a_{1} + a_{2} = 0 \dots (1)$
Putting $x = \frac{π}{4}$ , we get
$a_{1} + \frac{a_{3}}{2} = 0 \dots (2)$
From $(1)$ and $(2)$ , we get
$a_{2} = - a_{1}, a_{3} = - 2 a_{1}$

Hence, infinite triplets are possible.

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