Find the principal solution of cos 2x= sin x

$x = \frac{π}{6}$

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$x = \frac{π}{2}$

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$x = - \frac{π}{6}$

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$x = - \frac{π}{2}$

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Solution

The correct options are
A
$x = \frac{π}{6}$

D
$x = - \frac{π}{2}$

$c o s 2 x = s i n x 1 - 2 s i n^{2} x = s i n x 2 s i n^{2} x + s i n x - 1 = 0 L e t s i n x = t 2 t^{2} + t - 1 = 0 t = \frac{- 1 \pm \sqrt{1 - 4 \times 2 \times - 1}}{2 \times 2} t = \frac{- 1 \pm 3}{4} s i n x = \frac{1}{2} o r s i n x = - 1 x = \frac{π}{6} o r x = - \frac{π}{2}$

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