If x in left(0, frac{pi}{2}right) and x satis | vedclass

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

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

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(A) Given the equation $\sin x \cos x = \frac{1}{4}$.Multiply both sides by $2$: $2 \sin x \cos x = 2 	imes \frac{1}{4} = \frac{1}{2}$.Using the identity $\sin 2x = 2 \sin x \cos x$,we get $\sin 2x = \frac{1}{2}$.Since $\sin 2x = \sin \frac{\pi}{6}$,the general solution for $2x$ is $2x = n\pi + (-1)^n \frac{\pi}{6}$.For $n=0$,$2x = \frac{\pi}{6} \implies x = \frac{\pi}{12}$.For $n=1$,$2x = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \implies x = \frac{5\pi}{12}$.Both values $x = \frac{\pi}{12}$ and $x = \frac{5\pi}{12}$ lie in the interval $\left(0, \frac{\pi}{2}ight)$.

If $x \in \left(0, \frac{\pi}{2}\right)$ and $x$ satisfies the equation $\sin x \cos x = \frac{1}{4}$,then the values of $x$ are

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