Find the number of roots of the equation $3 sin x = 2 {cos}^{2} x$ in $[0, π]$

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Solution

Given, $3 sin x = 2 {cos}^{2} x$

$\Rightarrow 3 sin x = 2 (1 - {sin}^{2} x)$

$\Rightarrow 2 {sin}^{2} x + 3 sin x - 2 = 0$

$\Rightarrow 2 {sin}^{2} x + 4 sin x - sin x - 2 = 0$

$\Rightarrow 2 sin x (sin x + 2) - 1 (sin x + 2) = 0$

$\Rightarrow (sin x + 2) (2 sin x - 1) = 0$

$\Rightarrow sin x \neq - 2$ since range of sine function is $[- 1, 1]$

$∴ sin x = \frac{1}{2}$

$\Rightarrow x = \frac{π}{6}, π - \frac{π}{6}$

$∴ x = \frac{π}{6}, \frac{5 π}{6}$

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