If ${\log _{\cos x}}\sin x > 2\;{\rm{and}}\;0 < x < 3\pi ,\;$ then $\sin x\;$ lies in the interval

$\left[ {\frac{{\sqrt 5 - 1}}{2},1} \right]$

$\left[ {0,\frac{1}{2}} \right]$

JEE MATHEMATICS MCQ: FUNCTIONS

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Analyzing +4 | -1

{\log _{\cos x}}\sin x > 2\;\mathrm{and}\;0 < x < 3\pi ,\;

then

\sin x\;

lies in the interval

View answer and explanation

Correct option: B

\left[ {0,\frac{{\sqrt 5 - 1}}{2}} \right]

Option B

\left[ {0,\frac{{\sqrt 5 - 1}}{2}} \right]

is correct because it satisfies the logarithmic inequality and the domain restrictions for sine and cosine.