Question
The number of real Solutions of the equation $\sqrt {1 + \cos 2x} = \sqrt 2 {\cos ^{ - 1}}(\cos x){\mathop{\rm in}\nolimits} \left[ {\frac{\pi }{2},\pi } \right]$ is
- (a) 0 ✓ Correct
- (b) 1
- (c) 2
- (d) $\infty$
Inverse Trigonometric Functions — Class 12 Maths Solution
Step-by-step Solution
We have, $\sqrt {1 + \cos 2x} = \sqrt 2 {\cos ^{ - 1}}(\cos x),\left[ {\frac{\pi }{2},\pi } \right]$
$\Rightarrow$ $\sqrt {1 + 2{{\cos }^2}x - 1} = \sqrt 2 {\cos ^{ - 1}}(\cos x)$
$\Rightarrow$ $\sqrt 2 \cos x = \sqrt 2 {\cos ^{ - 1}}(\cos x)$
$\Rightarrow$ $\cos x = {\cos ^{ - 1}}(\cos x)$
$\Rightarrow$ $\cos x = x$
which is not true for any real value of $x$.
Hence, there is no Solution possible for the given equation.
NCERT & Exemplar solution for CBSE Class 12 Mathematics, Inverse Trigonometric Functions. Curated by Sachin Sharma. Free for all students.