Trigonometry & Inverse Trigonometry
Trigonometry
star_batch_jee_advanced_2025
Grade 11
Question:
In a $\triangle ABC$, $\angle A > \angle B$. Let $\angle A, \angle B$ satisfy the equation $3\sin x - 4\sin^3 x - k = 0$, where $0 < k < 1$, then $\angle C$ is equal to:
$\frac{\pi}{3}$
$\frac{\pi}{2}$
$\frac{2\pi}{3}$
None of these
Step-by-Step Solution
Key Concept: Use the triple angle formula $\sin 3x = 3\sin x - 4\sin^3 x$ to convert the equation into $\sin 3x = k$, then apply the sine equality condition.
From $3\sin x - 4\sin^3 x - k = 0$, we get $\sin 3x - k = 0$. Since $A$ and $B$ satisfy this equation, we have $\sin 3A = k$ and $\sin 3B = k$, which means $\sin 3A = \sin 3B$. This gives $3A = 180° - 3B$ (or $A - B$ possibility), so $3A + 3B = 180°$, yielding $A + B = 60°$. Therefore $C = 180° - 60° = 120°$ and expressing this in radians: $C = \frac{2\pi}{3}$.
Correct Answer: 3