Question

Solve ${\cos }^{-1}\sqrt{3}x+{\cos }^{-1}x=\frac{\mathrm{\pi }}{2}$

Answer 1

$$\begin{gathered} {\cos }^{-1}\sqrt{3}x+{\cos }^{-1}x=\frac{\mathrm{\pi }}{2} \\ \Rightarrow {\cos }^{-1}\left[\sqrt{3}x\times x-\sqrt{1-{\left(\sqrt{3}x\right)}^2}\sqrt{1-x^2}\right]=\frac{\mathrm{\pi }}{2}\left[\because {\cos }^{-1}x+{\cos }^{-1}y={\cos }^{-1}\left(xy-\sqrt{1-x^2}\sqrt{1-y^2}\right)\right] \\ \Rightarrow {\cos }^{-1}\left[\sqrt{3}{x}^2-\sqrt{1-3x^2}\sqrt{1-x^2}\right]=\frac{\mathrm{\pi }}{2} \\ \Rightarrow \sqrt{3}{x}^2-\sqrt{1-3x^2}\sqrt{1-x^2}=\cos \frac{\mathrm{\pi }}{2} \\ \Rightarrow \sqrt{3}{x}^2=\sqrt{1-3x^2}\sqrt{1-x^2} \\ \Rightarrow 3x^4=\left(1-3x^2\right)\left(1-x^2\right) \end{gathered}$$
$$\begin{gathered} \Rightarrow 3x^4=1-3x^2+3x^4-x^2 \\ \Rightarrow 4x^2=1 \\ \Rightarrow x^2=\frac{1}{4} \\ \Rightarrow x=\pm \frac{1}{2} \end{gathered}$$