Consider the trigonometric equation $tan x (sin 2 x + 1) = sin x (2 + tan x)$ . The number of solution(s) of the equation in $(0, 4 π)$ is

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Solution

$tan x (sin 2 x + 1) = sin x (2 + tan x)$
$\Rightarrow sin x [\frac{sin 2 x + 1}{cos x} - (2 + tan x)] = 0$
$\Rightarrow sin x = 0$
or $sin 2 x + 1 = 2 cos x + sin x, cos x \neq 0$
$\Rightarrow sin 2 x - sin x + 1 - 2 cos x = 0$
$\Rightarrow sin x (2 cos x - 1) - 1 (2 cos x - 1) = 0$
$\Rightarrow (sin x - 1) (2 cos x - 1) = 0$
$\Rightarrow sin x = 1, cos x = \frac{1}{2}$

$sin x = 0, 1; cos x = \frac{1}{2}$
But $sin x \neq 1$
$∴ x = n π, 2 n π \pm \frac{π}{3}$
$x = π, 2 π, 3 π, \frac{π}{3}, 2 π \pm \frac{π}{3}, 4 π - \frac{π}{3}$
i.e., a total of $7$ solutions in $(0, 4 π)$

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