Given equation

$\log_{\frac12}|\sin x| = 2 - \log_{\frac12}|\cos x|$

We need the number of distinct solutions in $[0,2\pi]$.

Bring logarithms together

Move the logarithmic term on the right to the left:

$\log_{\frac12}|\sin x| + \log_{\frac12}|\cos x| = 2$

Using $\log_a m + \log_a n = \log_a(mn),$ we get

$\log_{\frac12}(|\sin x|\,|\cos x|)=2$

Hence,

$|\sin x|\,|\cos x| = \left(\frac12\right)^2 = \frac14$

Simplify using double-angle identity

Since $2|\sin x||\cos x| = |\sin 2x|,$ we get

$|\sin 2x| = 2\cdot \frac14 = \frac12$

So we must solve

$|\sin 2x| = \frac12$

Solve the trigonometric equation

The equation $|\sin\theta|=\frac12$ has solutions in one full cycle $[0,2\pi]$:

$\theta = \frac\pi6,\ \frac{5\pi}6,\ \frac{7\pi}6,\ \frac{11\pi}6$

Here $\theta=2x$. Since $x\in[0,2\pi]$, we have $2x\in[0,4\pi]$, i.e. two full cycles.

Therefore the number of solutions for $2x$ is $4\times 2 = 8$.

Corresponding values of $x$ are:

$x=\frac\pi{12},\ \frac{5\pi}{12},\ \frac{7\pi}{12},\ \frac{11\pi}{12},\ \frac{13\pi}{12},\ \frac{17\pi}{12},\ \frac{19\pi}{12},\ \frac{23\pi}{12}$

Check domain restrictions

Since logarithms require $|\sin x|>0 \quad \text{and} \quad |\cos x|>0,$ we must exclude points where $\sin x=0$ or $\cos x=0$.

None of the 8 values above make $\sin x$ or $\cos x$ zero, so all are valid.

Conclusion

The number of distinct solutions is

$\boxed{8}$