#### Answer

The solution set is $$\{\frac{\pi}{12},\frac{5\pi}{12},\frac{13\pi}{12},\frac{17\pi}{12}\}$$

#### Work Step by Step

$$\sin x\cos x=\frac{1}{4}$$ over interval $[0,2\pi)$
This equation features both sine and cosine of the same functions of $x$, and they are in the position of a multiplication. In fact, it recalls the identity:
$$2\sin x\cos x=\sin 2x$$
$$\sin x\cos x=\frac{\sin 2x}{2}$$
Therefore, the given equation would now become
$$\frac{\sin2x}{2}=\frac{1}{4}$$
$$\sin2x=\frac{1}{2}$$
Using the identity $2\sin x\cos x=\sin2x$ helps solve this exercise, since if we keep both sine and cosine, there is no known way to solve it (except probably graphing).
1) Find corresponding interval for $2x$:
The interval for $x$ is $[0,2\pi)$. In other words,
$$0\le x\lt2\pi$$
Thus, with $2x$, the inequality would be
$$0\le2x\lt4\pi$$
So the interval for $2x$ is $[0,4\pi)$
2) Now consider back the equation $$\sin2x=\frac{1}{2}$$
Over interval $[0,4\pi)$, there are 4 values whose $\sin$ equals $\frac{1}{2}$, which are $\{\frac{\pi}{6},\frac{5\pi}{6},\frac{13\pi}{6},\frac{17\pi}{6}\}$
Therefore, $$2x\in\{\frac{\pi}{6},\frac{5\pi}{6},\frac{13\pi}{6},\frac{17\pi}{6}\}$$
That means
$$x\in\{\frac{\pi}{12},\frac{5\pi}{12},\frac{13\pi}{12},\frac{17\pi}{12}\}$$