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The number of distinct solutions of the equation $\log _{\frac{1}{2}}|\sin x|=2-\log _{\frac{1}{2}}|\cos x|$ in the interval $[0,2 \pi],$ is
A
$8$
B
$5$
C
$11$
D
$12$
(JEE MAIN-2020)
Solution
$\log _{1 / 2}|\sin x|=2-\log _{1 / 2}|\cos x| ; x \in[0,2 \pi]$
$\Rightarrow \quad \log _{1 / 2}|\sin x|+\log _{1 / 2}|\cos x|=2$
$\Rightarrow \log _{1 / 2}(|\sin x \cos x|)=2$
$\Rightarrow \quad|\sin x \cos x|=\frac{1}{4} \quad \Rightarrow|\sin 2 x|=\frac{1}{2}$
$=8$ Solutions
Standard 11
Mathematics
