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

Hint: (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.