Trigonometry & Inverse Trigonometry
Solution set and sum of inverse trig equation
nta_pyq_2023_jan
Grade None
Question:
Let S be the set of all solutions of the equation \cos^{-1}(2x) - 2\cos^{-1}(\sqrt{1-x^2}) = \pi, \quad x \in \left[-\frac{1}{2}, \frac{1}{2}\right]. Then \sum_{x \in S} 2\sin^{-1}(x^2 - 1) \text{ is equal to}
0
\frac{-2\pi}{3}
\pi - \sin^{-1}\left(\frac{\sqrt{3}}{4}\right)
\pi - 2\sin^{-1}\left(\frac{\sqrt{3}}{4}\right)
Step-by-Step Solution
Key Concept: Use the identity \cos^{-1}(\sqrt{1-x^2}) = \sin^{-1}|x| to simplify; solve quadratic for x.
Equation reduces to 2x^2 - 2x - 1 = 0. Valid solution: x = (1-\sqrt{3})/2. Then 2\sin^{-1}(x^2-1) = 2\sin^{-1}(-\sqrt{3}/2) = -2\pi/3.
Correct Answer: 2