If $4 {cos}^{2} x sin x - 2 {sin}^{2} x = 3 sin x$ , then find the value of x.

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Solution

$4 {cos}^{2} x sin x - 2 {sin}^{2} x = 3 sin x$
$4 (1 - {sin}^{2} x) sin x - 2 {sin}^{2} x = 3 sin x$
$4 sin x - 4 {sin}^{3} x - 2 {sin}^{2} x = 3 sin x$
$- 4 {sin}^{3} x - 2 {sin}^{2} x + sin x = 0$
$sin x (- 4 {sin}^{2} x - 2 sin x + 1) = 0$
$sin x = 0 o r - 4 {sin}^{2} x - 2 sin x + 1 = 0$
$\begin{matrix} - 4 {sin}^{2} x - 2 sin x + 1 = 0 sin x & = \frac{2 \pm \sqrt{4 + 16}}{- \frac{8}{1}} = \frac{2 \pm 2 \sqrt{5}}{- 8} \end{matrix}$
$= \frac{\pm \sqrt{5} + 1}{- 4}$

$= (\frac{1 + \sqrt{5}}{4}) & - (\frac{1 - \sqrt{5}}{4})$
$\begin{matrix} sin x = & - (\frac{1 + \sqrt{5}}{4}) = sin (- \frac{3 π}{10}) \end{matrix}$

or, $\begin{matrix} sin x & = \frac{- 1 + \sqrt{5}}{4} = sin \frac{π}{10} \end{matrix}$
General solution:
$x = n π \to 0$
$x = n π + (- 1)^{\frac{π}{10}}$
$x = n π + (- 1)^{n} (\frac{- 3 π}{10})$

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