1. Let $z=\frac{1+i\cos\theta}{1-2i\cos\theta}.$ We want $z$ to be purely imaginary, so its real part must be zero.

2. Put $x=\cos\theta.$ Then $z=\frac{1+ix}{1-2ix}.$

3. Rationalize the denominator: $z=\frac{(1+ix)(1+2ix)}{(1-2ix)(1+2ix)}.$ Now, $(1+ix)(1+2ix)=1+2ix+ix+2i^2x^2=1+3ix-2x^2,$ since $i^2=-1$. Also, $(1-2ix)(1+2ix)=1+4x^2.$ Hence, $z=\frac{1-2x^2+3ix}{1+4x^2}.$

4. Therefore, $\Re(z)=\frac{1-2x^2}{1+4x^2}.$ For $z$ to be purely imaginary, $\Re(z)=0 \implies 1-2x^2=0.$ So, $x^2=\frac12 \implies x=\pm \frac{1}{\sqrt2}.$ Thus, $\cos\theta=\pm \frac{1}{\sqrt2}.$

5. Now find all $\theta\in[-\pi,2\pi]$ satisfying this.

For $\cos\theta=\frac{1}{\sqrt2}$: $\theta=2k\pi\pm \frac{\pi}{4}.$ Within $[-\pi,2\pi]$, these are $-\frac{\pi}{4},\ \frac{\pi}{4},\ \frac{7\pi}{4}.$

For $\cos\theta=-\frac{1}{\sqrt2}$: $\theta=2k\pi\pm \frac{3\pi}{4}.$ Within $[-\pi,2\pi]$, these are $-\frac{3\pi}{4},\ \frac{3\pi}{4},\ \frac{5\pi}{4}.$

So all possible values are $-\frac{3\pi}{4},\ -\frac{\pi}{4},\ \frac{\pi}{4},\ \frac{3\pi}{4},\ \frac{5\pi}{4},\ \frac{7\pi}{4}.$

6. Their sum is $\left(-\frac{3\pi}{4}-\frac{\pi}{4}+\frac{\pi}{4}+\frac{3\pi}{4}\right)+\left(\frac{5\pi}{4}+\frac{7\pi}{4}\right).$ The first four terms cancel to $0$, so $\text{sum}=\frac{12\pi}{4}=3\pi.$

7. Therefore, the correct option is $\boxed{3\pi}.$ So option B is correct.