Question

The number of real solutions of the equation
$\sqrt{1+\cos 2x}=\sqrt{2}{\sin }^{-1}(\sin x),-\mathrm{\pi }\le x\mathit{\le }\mathrm{\pi }$ is
(a) 0
(b) 1
(c) 2
(d) infinite

Answer 1

(c) 2

$$\begin{gathered} \mathrm{For},-\mathrm{\pi }\le x\mathit{\le }\frac{-\mathrm{\pi }}{2} \\ \sqrt{1+\cos 2x}=\sqrt{2}{\sin }^{-1}(\sin x) \\ \Rightarrow \sqrt{2}\left|\cos x\right|=\sqrt{2}\left(-\mathrm{\pi }-\mathrm{x}\right) \\ \Rightarrow \sqrt{2}\left(-\cos x\right)=\sqrt{2}\left(-\mathrm{\pi }-\mathrm{x}\right) \\ \Rightarrow \cos x=\mathrm{\pi }+\mathrm{x} \\ \mathrm{It}\mathrm{does}\mathrm{not}\mathrm{satisfy}\mathrm{for}\mathrm{any}\mathrm{value}\mathrm{of}x\mathrm{in}\mathrm{the}\mathrm{interval}\left(-\mathrm{\pi },\frac{-\mathrm{\pi }}{2}\right) \end{gathered}$$

$$\begin{gathered} \mathrm{For},\frac{-\mathrm{\pi }}{2}\le x\mathit{\le }\frac{\mathrm{\pi }}{2} \\ \sqrt{1+\cos 2x}=\sqrt{2}{\sin }^{-1}(\sin x) \\ \Rightarrow \sqrt{2}\left|\cos x\right|=\sqrt{2}\left(\mathrm{x}\right) \\ \Rightarrow \sqrt{2}\left(\cos x\right)=\sqrt{2}\left(\mathrm{x}\right) \\ \Rightarrow \cos x=\mathrm{x} \\ \mathrm{It}\mathrm{gives}\mathrm{one}\mathrm{value}\mathrm{of}x\mathrm{in}\mathrm{the}\mathrm{interval}\left(\frac{-\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right) \end{gathered}$$

$$\begin{gathered} \mathrm{For},\frac{\mathrm{\pi }}{2}\le x\mathit{\le }\mathrm{\pi } \\ \sqrt{1+\cos 2x}=\sqrt{2}{\sin }^{-1}(\sin x) \\ \Rightarrow \sqrt{2}\left|\cos x\right|=\sqrt{2}\left(-\mathrm{\pi }-\mathrm{x}\right) \\ \Rightarrow \sqrt{2}\left(-\cos x\right)=\sqrt{2}\left(\mathrm{\pi }-\mathrm{x}\right) \\ \Rightarrow \cos x=-\mathrm{\pi }+\mathrm{x} \\ \mathrm{It}\mathrm{gives}\mathrm{one}\mathrm{value}\mathrm{of}x\mathrm{in}\mathrm{the}\mathrm{interval}\left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right) \end{gathered}$$

$\therefore \sqrt{1+\cos 2x}=\sqrt{2}{\sin }^{-1}(\sin x)\mathrm{gives}\mathrm{two}\mathrm{real}\mathrm{solutions}\mathrm{in}\mathrm{the}\mathrm{interval}\left[-\mathrm{\pi }\mathit{,}\mathit{}\mathrm{\pi }\right]$