If ${log}_{0.5} sin x = 1 - {log}_{0.5} cos x$ , then find number of value of $x ϵ [- 2 π, 2 π]$

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Solution

We have,
${log}_{0.5} sin x = 1 - {log}_{0.5} cos x$
${log}_{0.5} sin x cos x = 1 {log}_{2} sin x cos x = - 1 sin x cos x = \frac{1}{2} sin 2 x = 1 2 x = 2 n π + \frac{π}{2} x = n π + \frac{π}{4}$

Where $n = - 2, - 1, 0, 1$

Therefore,
$x = \frac{- 7 π}{4}, \frac{- 73 π}{4}, \frac{π}{4}, \frac{55 π}{4}$

$B u t f o r \frac{5 π}{4} sin x < 0 & cos x < 0$
$F o r \frac{- 3 π}{4} sin x < 0 x = \frac{- 7 π}{4}$
And $\frac{π}{4}$

Hence, this is the answer.

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