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: Testing the boundary value x=0 first:
Inverse-sine equations of this type, valid only on a restricted domain, are often solved efficiently by testing simple candidate values first: at \(x=0\), LHS \(=\sin^{-1}(1)-2\sin^{-1}(0)=\pi/2-0=\pi/2\), which matches the RHS exactly.

Step 2: Confirming uniqueness algebraically:
Squaring/rearranging the sine-of-both-sides equation \(1-x=1-2x^2\) gives \(x(2x-1)=0\), so \(x=0\) or \(x=1/2\); substituting \(x=1/2\) back into the ORIGINAL equation fails (gives \(-\pi/6\)), confirming it is extraneous.

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