Trigonometry & Inverse Trigonometry
Trigonometry
star_batch_jee_advanced_2025
Grade 11
Question:
If the equation $a_1 + a_2 \cos 2x + a_3 \sin^2 x = 1$ is satisfied by every real value of $x$, then the number of possible values of the triplet $(a_1, a_2, a_3)$ is:
Step-by-Step Solution
Key Concept: Demanding an identity hold for all values of the variable forces all coefficients of independent terms to zero.
Given $a_1 + a_2\cos 2x + a_3\sin^2 x = 1$ for all $x$, rewrite as $a_1 + a_2(1-2\sin^2 x) + a_3\sin^2 x = 1$, which yields $\sin^2 x(a_3 - 2a_2) + (a_1 + a_2 - 1) = 0$. For this to hold for all $x$, both coefficients must vanish: $a_3 - 2a_2 = 0$ and $a_1 + a_2 - 1 = 0$, giving three unknowns with two equations—infinitely many solutions.
Correct Answer: 4