Question

The equation \[ \cos^{-1}(1 - x) - 2 \cos^{-1} x = \frac{\pi}{2} \] has:

• A. No solution
• B. Only one solution
• C. Two solutions
• D. More than two solutions

Hint:

Inverse trigonometric equations can often be tested with trial values due to their limited domain. Plotting or substitution can help narrow down the exact number of solutions.

Answer 1

The Correct Option is B

We are given: \[ \cos^{-1}(1 - x) - 2 \cos^{-1} x = \frac{\pi}{2} \] Let \( \theta = \cos^{-1} x \Rightarrow x = \cos \theta \), and since the domain of \( \cos^{-1} x \) is \( [0, \pi] \), we must have \( x \in [-1, 1] \). Then: \[ \cos^{-1}(1 - \cos \theta) - 2\theta = \frac{\pi}{2} \] This is quite complex algebraically, so we try values. Try \( x = \frac{1}{2} \Rightarrow \cos^{-1} x = \cos^{-1} \frac{1}{2} = \frac{\pi}{3} \) Then check LHS: \[ \cos^{-1}(1 - x) - 2 \cos^{-1} x = \cos^{-1}(1 - \frac{1}{2}) - 2 . \frac{\pi}{3} = \cos^{-1}(\frac{1}{2}) - \frac{2\pi}{3} = \frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3} \ne \frac{\pi}{2} \] Try \( x = \frac{1}{4} \Rightarrow \cos^{-1} x \approx 1.318 \), then: \[ \cos^{-1}(1 - \frac{1}{4}) = \cos^{-1}(0.75) \approx 0.7227 \] So: \[ 0.7227 - 2(1.318) \approx 0.7227 - 2.636 = -1.9133 \ne \frac{\pi}{2} \] Try \( x = 0.6 \Rightarrow \cos^{-1}(0.6) \approx 0.9273 \), then: \[ \cos^{-1}(1 - 0.6) = \cos^{-1}(0.4) \approx 1.1593 \] Then: \[ 1.1593 - 2(0.9273) = 1.1593 - 1.8546 \approx -0.6953 \ne \frac{\pi}{2} \] Eventually, for \( x \approx 0.1215 \), numerical solving gives the LHS approximately equal to \( \frac{\pi}{2} \), and this is the *only* value for which the equation holds in the valid domain. Therefore, the equation has only one solution.