If sin x cos x = frac{1}{4}, then the general | vedclass

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

Vedclass · Accepted Answer

$\frac{n\pi}{2} + (-1)^n \frac{\pi}{12}$

Vedclass · Answer

(A) Given the equation $\sin x \cos x = \frac{1}{4}$.
Multiply both sides by $2$ to use the identity $\sin(2x) = 2 \sin x \cos x$:
$2 \sin x \cos x = 2 	imes \frac{1}{4}$
$\sin(2x) = \frac{1}{2}$
We know that $\sin(\frac{\pi}{6}) = \frac{1}{2}$.
The general solution for $\sin 	heta = \sin \alpha$ is $	heta = n\pi + (-1)^n \alpha$.
Here, $	heta = 2x$ and $\alpha = \frac{\pi}{6}$.
So, $2x = n\pi + (-1)^n \frac{\pi}{6}$.
Dividing by $2$, we get:
$x = \frac{n\pi}{2} + (-1)^n \frac{\pi}{12}$.

If $\sin x \cos x = \frac{1}{4}$, then the general solution is:

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