Question

The total number of solutions of $\cos x = \sqrt {1 - \sin 2x}$ in $[0,2\pi ]$is equal to
A. 2
B. 3
C. 5
D. none of these

Answer 1

(i) Here, we are going to use the concept of modulus function.
(ii) The concept of sign change according to the quadrant.

Complete step-by-step answer:
Given, $\cos x = \sqrt {1 - \sin 2x}$
Consider $\sqrt {1 - \sin 2x} = \sqrt {{{\sin }^2}x + {{\cos }^2}x - 2\sin x\cos x}$

$$= \sqrt {{{(\sin x - \cos x)}^2}} \\\ = \left| {\sin x - \cos x} \right| \\\$$

Now, there are two cases either $\sin x > \cos x$ or $\cos x > \sin x$
Consider the case $\sin x > \cos x$ then if we observe the graph of $\sin x$ and $\cos x$ simultaneously
We observe that $\sin x > \cos x$ in the region $\left( {\dfrac{\pi }{4},\dfrac{{5\pi }}{4}} \right)$
And $\cos x > \sin x$ is in the region $\left( {0,\dfrac{\pi }{4}} \right) \cup \left( {\dfrac{{5\pi }}{4},2\pi } \right)$
Also, for the case $\sin x > \cos x$ we have $\cos x = \sin x - \cos x$ as given in the question

$$\Rightarrow 2\cos x = \sin x \\\ \Rightarrow \tan x = 2 \\\$$

Now, we know the principal value of $\tan x$ is given by $\tan x + n\pi ,n \in \\{ 1,2,....\\}$
Since, the interval to be considered is given to be $[0,2\pi ]$
Therefore, $x = {\tan ^{ - 1}}2,\pi + {\tan ^{ - 1}}2$
But since $\pi + {\tan ^{ - 1}}2$does not belong in the interval $\left( {\dfrac{\pi }{4},\dfrac{{5\pi }}{4}} \right)$ therefore it is not considered.
Hence, we have only one solution for the case $\sin x > \cos x$
Similarly, for the case $\cos x > \sin x$we have $\cos x = \cos x - \sin x$

$$\Rightarrow \sin x = 0 \\\ \Rightarrow x = n\pi ,n \in \\{ 0,1,2,....\\} \\\$$

Since, the interval to be considered is given to be$[0,2\pi ]$
Therefore, $x = 0,\pi ,2\pi$
But since \pi $$$$ \notin \left( {0,\dfrac{\pi }{4}} \right) \cup \left( {\dfrac{{5\pi }}{4},2\pi } \right)
Therefore, we have only two solutions for this case
Hence, total number of solutions is equal to three
And therefore, option B. 3 is the required answer

Note: (i) One should carefully look at the regions while considering the cases
(ii) One can also check if $\sin x$is greater than or less than $\cos x$by taking any particular value in the considered interval.