The number of all 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 is

0

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Solution

The correct option is D $\infty$
Given- $a_{1} + a_{2} cos 2 x + a_{3} {sin}^{2} . x = 0$ .

To find out- the number of the triplets $(a_{1}, a_{2}, a_{3})$ which will make the given mathematical sentence true.

Solution- $a_{1} + a_{2} cos 2 x + a_{3} {sin}^{2} . x = 0$

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

$\Rightarrow a_{1} + a_{2} + a_{3} - {2 a}_{2} {sin}^{2} x = 0$

$\Rightarrow {2 a}_{2} {sin}^{2} x = a_{1} + a_{2} + a_{3}$

$\Rightarrow {sin}^{2} x = \frac{a_{1} + a_{2} + a_{3}}{{2 a}_{2}}$ ..........(i)

Now for all values of $x 0 \leq {sin}^{2} x \leq 1$ and there are infinite number of points in between this interval. So the number of possible triplets $(a_{1}, a_{2}, a_{3})$ satisfying (i) is infinite.

Ans- Option D.

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