Find number of roots of the equation $sin x = 3 cos 2 x - 1$ in $[- π, π]$

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Solution

$sin x = 2 cos 2 x - 1$

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

$\Rightarrow sin x = 3 - 6 {sin}^{2} x - 1$

$\Rightarrow 6 {sin}^{2} x + sin x - 2 = 0$

$\Rightarrow 6 {sin}^{2} x + 4 sin x - 3 sin x - 2 = 0$

$\Rightarrow 2 sin x (3 sin x + 2) - 1 (3 sin x + 2) = 0$

$\Rightarrow (3 sin x + 2) (2 sin x - 1) = 0$

$\Rightarrow sin x = \frac{- 2}{3}, \frac{1}{2}$

$\Rightarrow sin x = \frac{1}{2}$

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

$x = \frac{π}{6}, \frac{5 π}{6}$

$\Rightarrow sin x = \frac{- 2}{3}$

$\Rightarrow x = - {sin}^{- 1} (\frac{2}{3}), - π + {sin}^{- 1} (\frac{2}{3})$

$∴ x = \frac{π}{6}, π - \frac{π}{6}, - {sin}^{- 1} (\frac{2}{3}), - π + {sin}^{- 1} (\frac{2}{3})$

Thus, there are $4$ solutions.

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