The number of solution(s) of $2 sin 2 x + sin x (1 - 2 cos 2 x) - 4 = 0$ , where $x \in (0, π$ ), is

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Solution

Given equation is
$2 sin 2 x + sin x (1 - 2 cos 2 x) - 4 = 0 \Rightarrow 4 sin x cos x + sin x (1 - 2 cos 2 x) - 4 = 0 \Rightarrow 4 cos x + 1 - 2 cos 2 x = \frac{4}{sin x} \Rightarrow 4 cos x + 1 - 2 (2 {cos}^{2} x - 1) = \frac{4}{sin x} \Rightarrow - 4 {cos}^{2} x + 4 cos x + 3 = \frac{4}{sin x} \Rightarrow 4 - (2 cos x - 1)^{2} = \frac{4}{sin x}$

$L . H . S \leq 4 R . H . S \geq 4$
They will be same when both is equal to $4$ , so
$sin x = 1 \Rightarrow x = \frac{π}{2} cos x = \frac{1}{2} \Rightarrow x = \frac{π}{3}$

Hence, there is no solution.

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