Complete set of values of x satisfying cos 2 x-sin x- x -2- - isA- -6- 5 -6B- -6- - -6- -6 -5 -6- -D- -6- -2 -5 -6- -

The correct option is C ( π/6, π/6 ) ∪ ( 5 π/6, π ) If cos2x=sinx ⇒ 2 sin^2x+sinx 1=0 ∴ sinx=1/2, 1 But sinx ≠ 1 [as cos2x gt;1 is incorrect] ∴ sinx=1/2 ...

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Standard XII

Mathematics

Trigonometric Equations

Complete set ...

Question

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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The correct option is B (−π6,π6)∪(5π6,π) If cos2x=sinx ⇒2sin2x+sinx−1=0∴sinx=12,−1 But sinx≠−1 [as cos2x>1 is incorrect] ∴sinx=12 ⇒x=π6,5π6. ∴ Solution set is (−π6,π6)∪(5π6,π)

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