The number of distinct solutions of the equat | vedclass

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(A) Given equation: $\log _{1 / 2}|\sin x|=2-\log _{1 / 2}|\cos x|$ for $x \in [0, 2\pi]$.Rearranging the terms,we get: $\log _{1 / 2}|\sin x| + \log

Vedclass · Accepted Answer

$8$

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(A) Given equation: $\log _{1 / 2}|\sin x|=2-\log _{1 / 2}|\cos x|$ for $x \in [0, 2\pi]$.Rearranging the terms,we get: $\log _{1 / 2}|\sin x| + \log _{1 / 2}|\cos x| = 2$.Using the property $\log_a m + \log_a n = \log_a (mn)$,we have: $\log _{1 / 2}(|\sin x \cos x|) = 2$.Converting to exponential form: $|\sin x \cos x| = (1/2)^2 = 1/4$.Multiplying by $2$ on both sides: $|2 \sin x \cos x| = 2 	imes (1/4) = 1/2$.Thus,$|\sin 2x| = 1/2$.In the interval $x \in [0, 2\pi]$,the angle $2x$ lies in the interval $[0, 4\pi]$.For $|\sin 	heta| = 1/2$,there are $4$ solutions in each interval of length $2\pi$ (i.e.,$[0, 2\pi]$).Since the interval for $2x$ is $[0, 4\pi]$,which covers two full periods of the sine function,the total number of solutions is $4 	imes 2 = 8$.

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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