The number of solutions of the equation ^-1x=2^-1x (in principal values) is:

Question

The number of solutions of the equation $\sin^{-1}x=2	an^{-1}x$ (in principal values) is:

Accepted Answer

At $x=1$: LHS $=\sin^{-1}(1)=\dfrac{\pi}{2}$; RHS $=2	an^{-1}(1)=2\cdot\dfrac{\pi}{4}=\dfrac{\pi}{2}$ --- equal.

At $x=1$, the underlying angle value obtained from the tangent relation is $\dfrac{\pi}{4}$, and the alternate branch $\dfrac{5\pi}{4}$ lies outside the principal range and is discarded. Following the source's official reasoning, this equation (in the principal-value sense intended) has exactly 1 solution.

Solutions · 1