Question

The number of real solutions of the equation \[\sqrt{1 + \cos 2x} = \sqrt{2} \sin^{- 1} (\sin x), - \pi \leq x \leq \pi\]

विकल्प

• 0

• 1

• 2

• infinite

Answer 1

(c) 2

\[For, - \pi \leq x \leq \frac{- \pi}{2}\]
\[\sqrt{1 + \cos 2x} = \sqrt{2} \sin^{- 1} (\sin x)\]
\[ \Rightarrow \sqrt{2} \left| \cos x \right| = \sqrt{2} \left( - \pi - x \right)\]
\[ \Rightarrow \sqrt{2} \left( - \cos x \right) = \sqrt{2} \left( - \pi - x \right)\]
\[ \Rightarrow \cos{x} = \pi + x \]
\[\text{ It does not satisfy for any value of x in the interval }\left( - \pi, \frac{- \pi}{2} \right)\]
\[For, \frac{- \pi}{2} \leq x \leq \frac{\pi}{2}\]
\[\sqrt{1 + \cos 2x} = \sqrt{2} \sin^{- 1} (\sin x)\]
\[ \Rightarrow \sqrt{2} \left| \cos x \right| = \sqrt{2} \left( x \right)\]
\[ \Rightarrow \sqrt{2} \left( \cos x \right) = \sqrt{2} \left( x \right)\]
\[ \Rightarrow \cos{x} = x \]
\[\text{ It gives one value of x in the interval }\left( \frac{- \pi}{2}, \frac{\pi}{2} \right)\]
\[For, \frac{\pi}{2} \leq x \leq \pi\]
\[\sqrt{1 + \cos 2x} = \sqrt{2} \sin^{- 1} (\sin x)\]
\[ \Rightarrow \sqrt{2} \left| \cos x \right| = \sqrt{2} \left( - \pi - x \right)\]
\[ \Rightarrow \sqrt{2} \left( - \cos x \right) = \sqrt{2} \left( \pi - x \right)\]
\[ \Rightarrow \cos{x} = - \pi + x \]
\[\text{ It gives one value of x in the interval } \left( \frac{\pi}{2}, \pi \right)\]
\[\therefore \sqrt{1 + \cos 2x} = \sqrt{2} \sin^{- 1} (\sin x) \text {gives two real solutions in the interval }\left[ - \pi, \pi \right]\]