The values of x in left(0, frac{pi}{2}right) | vedclass

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(B) Given the equation: $\sin x \cos x = \frac{1}{4}$.Multiply both sides by $2$: $2 \sin x \cos x = \frac{2}{4} = \frac{1}{2}$.Using the identity $\s

Vedclass · Accepted Answer

$\frac{\pi}{12}, \frac{5 \pi}{12}$

Vedclass · Answer

(B) Given the equation: $\sin x \cos x = \frac{1}{4}$.Multiply both sides by $2$: $2 \sin x \cos x = \frac{2}{4} = \frac{1}{2}$.Using the identity $\sin 2x = 2 \sin x \cos x$,we get $\sin 2x = \frac{1}{2}$.Since $x \in \left(0, \frac{\pi}{2}\right)$,then $2x \in (0, \pi)$.The solutions for $2x$ in $(0, \pi)$ are $2x = \frac{\pi}{6}$ and $2x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.Therefore,$x = \frac{\pi}{12}$ and $x = \frac{5\pi}{12}$.

The values of $x$ in $\left(0, \frac{\pi}{2}\right)$ satisfying the equation $\sin x \cos x = \frac{1}{4}$ are $.......$

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