Complete set of values of x satisfying cos2x > |sinx|, $x \in (\frac{- π}{2}, π)$ is

$(\frac{- π}{6}, π)$

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$(\frac{- π}{6}, \frac{π}{6}) \cup (\frac{5 π}{6}, π)$

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$(\frac{π}{6}, \frac{π}{2}) \cup (\frac{5 π}{6}, π)$

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$(\frac{- π}{6}, \frac{5 π}{6})$

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Solution

The correct option is B
$(\frac{- π}{6}, \frac{π}{6}) \cup (\frac{5 π}{6}, π)$

If $c o s 2 x = s i n x$
$\Rightarrow 2 s i n^{2} x + s i n x - 1 = 0 ∴ s i n x = \frac{1}{2}, - 1$
But $s i n x \neq - 1$ [as $c o s 2 x > 1$ is incorrect]
$∴ s i n x = \frac{1}{2} \Rightarrow x = \frac{π}{6}, \frac{5 π}{6} .$

$∴$ Solution set is
$(\frac{- π}{6}, \frac{π}{6}) \cup (\frac{5 π}{6}, π)$

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Similar questions

Complete set of values of x satisfying cos2x > |sinx|, $x \in (\frac{- π}{2}, π)$ is

x

sin x > \frac{1}{2}

S \in (0, π)

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1 + | cos x | + {cos}^{2} x + | {cos}^{3} x | + . . . \infty = 2

S =

x \in (- \frac{π}{2}, \frac{π}{2})

| 4 sin x + \sqrt{2} | < \sqrt{6}

(\frac{a π}{24}, \frac{b π}{24})

∣ ∣ ∣ \frac{a - b}{3} ∣ ∣ ∣ =

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