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Local Maxima

The solution ...

Question

The solution of the inequalities ${log}_{1 / 2} sin x > {log}_{1 / 2} cos x$ in $[0, 2 π]$ is

x \in (0, \frac{π}{2})

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x \in (0, \frac{π}{8})

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x \in (0, \frac{π}{4})

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None of these

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Solution

The correct option is C $x \in (0, \frac{π}{4})$
Hence the option C is correct.

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Similar questions

l o g_{\frac{1}{\sqrt{2}}} s i n x > 0, x ϵ [0, 4 π]

\frac{π}{4}

{log}_{\frac{1}{\sqrt{2}}} sin x \geq 0

x \in [0, 4 π]

x

\frac{π}{4}

{sin}^{5} x - {cos}^{5} x = \frac{1}{cos x} - \frac{1}{sin x}

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f (x) = \int \frac{x^{2}}{(1 + x^{2}) (1 + \sqrt{1 + x^{2}})} d x

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