The number of distinct solutions of the equat | vedclass

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Question

$\log _{1 / 2}|\sin x|=2-\log _{1 / 2}|\cos x| ; x \in[0,2 \pi]$

$\Rightarrow \quad \log _{1 / 2}|\sin x|+\log _{1 / 2}|\cos x|=2$

$\Rightarrow

Vedclass · Accepted Answer

$8$

Vedclass · Answer

$\log _{1 / 2}|\sin x|=2-\log _{1 / 2}|\cos x| ; x \in[0,2 \pi]$

$\Rightarrow \quad \log _{1 / 2}|\sin x|+\log _{1 / 2}|\cos x|=2$

$\Rightarrow \log _{1 / 2}(|\sin x \cos x|)=2$

$\Rightarrow \quad|\sin x \cos x|=\frac{1}{4} \quad \Rightarrow|\sin 2 x|=\frac{1}{2}$

$=8$ Solutions

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

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