The solution of the inequality ${\log _{1/2}}\sin x > {\rm{lo}}{{\rm{g}}_{1/2}}\cos x$ in $\left( {0,2\pi } \right)$ is

$x \in \left( {\frac{{5\pi }}{4},\,\;2\pi } \right)$

$x \in \left( {\;0,\,\,\frac{\pi }{4}} \right)\,\, \cup \,\,\left( {\frac{{5\pi }}{4},\;\,2\pi } \right)$

JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

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Analyzing +4 | -1

The solution of the inequality

{\log _{1/2}}\sin x > \mathrm{lo}{\mathrm{g}_{1/2}}\cos x

\left( {0,2\pi } \right)

View answer and explanation

Correct option: B

x \in \left( {0,\,\frac{\pi }{4}} \right)

Option B is correct because it finds the interval where both sin x > 0, cos x > 0, and sin x < cos x, which is (0, π/4).

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