The number of all the 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

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Solution

The correct option is D
infinite

Since $a_{1} + a_{2} c o s 2 x + a_{3} s i n^{2} x = 0 f i r a k k x$
Putting $x = 0 a n d x = \frac{π}{2}$ , we get
$a_{1} + a_{2} = 0 a n d a_{1} - a_{2} + a_{3} = 0 \Rightarrow a_{2} = - a_{1} a n d a_{3} = - 2 a_{1}$
Therefore, the given equation becomes
$a_{1} - a_{1} c o s 2 x - 2 a_{1} s i n^{2} x = 0, \forall x$
or $a_{1} (1 - c o s 2 x - 2 s i n^{2} x) = 0, \forall x$
or $a_{1} (2 s i n^{2} x - 2 s i n^{2} x) = 0, \forall x$
The above is satisfied for all values of $a_{1} .$
Hence, the infinite number of triplets $(a_{1}, - a_{1}, - 2 a_{1})$ is possible.
Alternative Method:
$a_{1} + a_{2} c o s 2 x + a_{3} s i n^{2} x = 0 f o r r e a l x ∴ a_{1} + a_{2} (1 - 2 s i n^{2} x) + a_{3} s i n^{2} x = 0 f o r r e a l x$
$∴ (a_{1} + a_{2}) + (a_{3} - 2 a_{2}) s i n^{2} x = 0$ for real x
$∴ a_{1} + a_{2} = 0 a n d a_{3} - 2 a_{2} = 0 ∴ a_{2} = - a_{1} a n d a_{3} = 2 a_{2} = - 2 a_{1}$
hence, infinite number of triplets $(a_{1}, - a_{1}, - 2 a_{1})$ exist.

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