Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

The equation is ${\sin }^{-1}(x)+{\sin }^{-1}(2x)=\frac{\pi }{3}$. For the inverse sine functions to be defined, we must have $-1\le x\le 1$ and $-1\le 2x\le 1$, which implies $x\in \left[-\frac{1}{2},\frac{1}{2}\right]$. Rearranging, ${\sin }^{-1}(x)=\frac{\pi }{3}-{\sin }^{-1}(2x)$. Taking sine on both sides:$$x=\sin \left(\frac{\pi }{3}-{\sin }^{-1}(2x)\right)=\sin \left(\frac{\pi }{3}\right)\cos \left({\sin }^{-1}(2x)\right)-\cos \left(\frac{\pi }{3}\right)\sin \left({\sin }^{-1}(2x)\right)$$ $$x=\frac{\sqrt{3}}{2}\sqrt{1-(2x{)}^2}-\frac{1}{2}(2x)$$ $$x=\frac{\sqrt{3}}{2}\sqrt{1-4x^2}-x$$ $$2x=\frac{\sqrt{3}}{2}\sqrt{1-4x^2}$$ For this equation to hold, $2x\ge 0$, so $x\ge 0$. Combining with the domain, $x\in \left[0,\frac{1}{2}\right]$. Squaring both sides: $(2x{)}^2={\left(\frac{\sqrt{3}}{2}\right)}^2(1-4x^2)$.$$4x^2=\frac{3}{4}(1-4x^2)$$ $$16x^2=3-12x^2$$ $$28x^2=3⟹x^2=\frac{3}{28}$$ $$x=\pm \sqrt{\frac{3}{28}}=\pm \frac{\sqrt{3}}{2\sqrt{7}}$$ Since $x\ge 0$, we take $x=\frac{\sqrt{3}}{2\sqrt{7}}$. This value is approximately $0.327$, which lies in $\left[0,\frac{1}{2}\right]$. Thus, there is exactly one real solution.