The number of solutions of the equation $$2x + 3\tan x = \pi$$, $$x \in [-2\pi, 2\pi] - \left\{\pm\frac{\pi}{2}, \pm\frac{3\pi}{2}\right\}$$ is

Let $$f(x)=2x+3\tan x-\pi$$ defined on $$[-2\pi,2\pi]\setminus\left\{\pm\frac{\pi}{2},\;\pm\frac{3\pi}{2}\right\}$$.

Derivative:
$$f'(x)=2+3\sec^{2}x.$$

Since $$\sec^{2}x\ge 1$$ for every real $$x$$, we have $$f'(x)=2+3\sec^{2}x\gt 0$$. Thus $$f(x)$$ is strictly increasing on every interval in which it is continuous. Consequently, in each such interval $$f(x)=0$$ can have at most one root.

The vertical asymptotes of $$\tan x$$ that lie inside the given domain are at $$x=-\frac{3\pi}{2},\;-\frac{\pi}{2},\;\frac{\pi}{2},\;\frac{3\pi}{2}.$$ These points divide the domain into five continuous intervals:

$$I_1:\;[-2\pi,\,-\frac{3\pi}{2}),\qquad I_2:\;(-\frac{3\pi}{2},\,-\frac{\pi}{2}),$$
$$I_3:\;(-\frac{\pi}{2},\,\frac{\pi}{2}),\qquad I_4:\;(\frac{\pi}{2},\,\frac{3\pi}{2}),\qquad I_5:\;(\frac{3\pi}{2},\,2\pi].$$

Case 1: $$x\in I_1=[-2\pi,-\frac{3\pi}{2})$$

$$f(-2\pi)=2(-2\pi)+3\tan(-2\pi)-\pi=-5\pi\lt 0.$$ As $$x\rightarrow\left(-\frac{3\pi}{2}\right)^{-},\;\tan x\rightarrow +\infty\;\Longrightarrow\;f(x)\rightarrow +\infty.$$ With $$f(x)$$ increasing, it must cross zero exactly once in $$I_1$$.

Case 2: $$x\in I_2=(-\frac{3\pi}{2},-\frac{\pi}{2})$$

As $$x\rightarrow\left(-\frac{3\pi}{2}\right)^{+},\;\tan x\rightarrow -\infty\;\Longrightarrow\;f(x)\rightarrow -\infty.$$ As $$x\rightarrow\left(-\frac{\pi}{2}\right)^{-},\;\tan x\rightarrow +\infty\;\Longrightarrow\;f(x)\rightarrow +\infty.$$ Again, by monotonicity, exactly one root lies in $$I_2$$.

Case 3: $$x\in I_3=(-\frac{\pi}{2},\frac{\pi}{2})$$

$$\displaystyle\lim_{x\to(-\frac{\pi}{2})^{+}}f(x)=-\infty,\qquad \lim_{x\to(\frac{\pi}{2})^{-}}f(x)=+\infty.$$ Hence there is exactly one root in $$I_3$$.

Case 4: $$x\in I_4=(\frac{\pi}{2},\frac{3\pi}{2})$$

$$\displaystyle\lim_{x\to(\frac{\pi}{2})^{+}}f(x)=-\infty,\qquad \lim_{x\to(\frac{3\pi}{2})^{-}}f(x)=+\infty.$$ Therefore one root exists in $$I_4$$.

Case 5: $$x\in I_5=(\frac{3\pi}{2},2\pi]$$

$$\displaystyle\lim_{x\to(\frac{3\pi}{2})^{+}}f(x)=-\infty.$$ At the right end, $$f(2\pi)=2(2\pi)+3\tan(2\pi)-\pi=4\pi-\pi=3\pi\gt 0.$$ Hence $$f(x)$$ rises from $$-\infty$$ to a positive value, giving one more root in $$I_5$$.

Counting the unique root in each of the five intervals, the total number of solutions in the prescribed domain is $$5$$.

Therefore, the correct choice is Option B.