Question

The number of solutions of the equation $\cos x=\sqrt{1-\sin 2x}$ where $x\in [0,2\pi ]$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$

$\cos x=\sqrt{1-\sin 2x}$
Now, we know that
$\sqrt{1-\sin 2x}=\sqrt{(\sin x-\cos x{)}^2}\Rightarrow \sqrt{1-\sin 2x}=|\sin x-\cos x|$

$|\sin x-\cos x|=\{\begin{array}{cc}\sin x-\cos x,& x\in [\frac{\pi }{4},\frac{5\pi }{4}]\\ \cos x-\sin x& x\in [0,\frac{\pi }{4})\cup (\frac{5\pi }{4},2\pi ]\end{array}$


Case $1:$ When $x\in [0,\frac{\pi }{4})\cup (\frac{5\pi }{4},2\pi ]$
$\cos x=\cos x-\sin x\Rightarrow \sin x=0\Rightarrow x=0,2\pi$

Case $2:$ When $x\in [\frac{\pi }{4},\frac{5\pi }{4}]$
$\cos x=\sin x-\cos x\Rightarrow 2\cos x=\sin x\Rightarrow \tan x=2$
Which has a solution in $x\in [\frac{\pi }{4},\frac{\pi }{2}]$

Hence, the required number of solutions are $3$.