Question

If \(\sin^{-1}(1-x)-2\sin^{-1}x=\dfrac{\pi}{2}\), then find the value of \(x\).

Hint:

Isolate one inverse sine term, apply sine to both sides using sin(pi/2+theta)=cos(theta), then verify roots (extraneous roots are common with inverse trig equations).

Answer 1

Step 1: Key Approach:
Rewrite as \(\sin^{-1}(1-x)=\dfrac{\pi}{2}+2\sin^{-1}x\), then take sine of both sides using \(\sin(\pi/2+\theta)=\cos\theta\).

Step 2: Applying sine to both sides:
\(1-x=\cos(2\sin^{-1}x)=1-2\sin^{2}(\sin^{-1}x)=1-2x^{2}\).

Step 3: Solving the resulting equation:
\(1-x=1-2x^{2}\ \Rightarrow\ 2x^{2}-x=0\ \Rightarrow\ x(2x-1)=0\ \Rightarrow\ x=0\text{ or }x=\dfrac12\).

Step 4: Verifying which root is valid:
For \(x=\tfrac12\): \(\sin^{-1}(1-\tfrac12)-2\sin^{-1}(\tfrac12)=\dfrac{\pi}{6}-2\cdot\dfrac{\pi}{6}=-\dfrac{\pi}{6}\neq\dfrac{\pi}{2}\), so \(x=\tfrac12\) is rejected. For \(x=0\): \(\sin^{-1}(1)-2\sin^{-1}(0)=\dfrac{\pi}{2}-0=\dfrac{\pi}{2}\) ✓.

Final Answer:
\[ \boxed{x=0} \]