The number of all possible 5-triplets $(a_{1}, a_{2}, a_{3}, a_{4}, a_{5})$ such that $a_{1} + a_{2} sin x + a_{3} cos x + a_{4} sin 2 x + a_{5} cos 2 x = 0$ holds for all $x$ is:

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infinite

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Solution

The correct option is B 1
since given expression is true for all $x$

taking $x = 0$
$a_{1} + a_{3} + a_{5} = 0 ⟶ (1)$

taking $x = \frac{π}{2}$
$a_{1} + a_{2} - a_{5} = 0 ⟶ (2)$

taking $x = π$
$a_{1} - a_{3} + a_{5} = 0 ⟶ (3)$

taking $x = - \frac{π}{2}$
$a_{1} - a_{2} - a_{5} = 0 ⟶ (4)$

using all these equation only solution is,
$a_{1} = a_{2} = a_{3} = a_{4} = a_{5} = 0$
Hence, option 'B' is correct.

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