1. Let $f(x)=2x+3\tan x-\pi$ We need the number of solutions of $f(x)=0$ for $x\in[-2\pi,2\pi]\setminus\left\{\pm\frac{\pi}{2},\pm\frac{3\pi}{2}\right\}.$

2. The interval is naturally split by the discontinuities of $\tan x$ into: $[-2\pi,-\tfrac{3\pi}{2}),\;(-\tfrac{3\pi}{2},-\tfrac{\pi}{2}),\;(-\tfrac{\pi}{2},\tfrac{\pi}{2}),\;(\tfrac{\pi}{2},\tfrac{3\pi}{2}),\;(\tfrac{3\pi}{2},2\pi].$

3. On each such interval, $f$ is continuous and $f'(x)=2+3\sec^2 x>0.$ So $f$ is strictly increasing on each interval. Therefore, on each interval there can be at most one solution.

4. Now check the sign/limit of $f$ on each interval.

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Interval 1: $[-2\pi,-\tfrac{3\pi}{2})$

At $x=-2\pi$, $f(-2\pi)=2(-2\pi)+3\tan(-2\pi)-\pi=-4\pi+0-\pi=-5\pi<0.$ As $x\to\left(-\tfrac{3\pi}{2}\right)^-$, $\tan x\to +\infty$, so $f(x)\to +\infty.$ Since $f$ is increasing and changes sign, there is exactly one solution here.

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Interval 2: $(-\tfrac{3\pi}{2},-\tfrac{\pi}{2})$

As $x\to\left(-\tfrac{3\pi}{2}\right)^+$, $\tan x\to -\infty$, so $f(x)\to -\infty.$ As $x\to\left(-\tfrac{\pi}{2}\right)^-$, $\tan x\to +\infty$, so $f(x)\to +\infty.$ Hence exactly one solution in this interval.

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Interval 3: $(-\tfrac{\pi}{2},\tfrac{\pi}{2})$

As $x\to\left(-\tfrac{\pi}{2}\right)^+$, $\tan x\to -\infty$, so $f(x)\to -\infty.$ As $x\to\left(\tfrac{\pi}{2}\right)^-$, $\tan x\to +\infty$, so $f(x)\to +\infty.$ Hence exactly one solution in this interval.

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Interval 4: $(\tfrac{\pi}{2},\tfrac{3\pi}{2})$

As $x\to\left(\tfrac{\pi}{2}\right)^+$, $\tan x\to -\infty$, so $f(x)\to -\infty.$ As $x\to\left(\tfrac{3\pi}{2}\right)^-$, $\tan x\to +\infty,$ so $f(x)\to +\infty.$ Hence exactly one solution in this interval.

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Interval 5: $(\tfrac{3\pi}{2},2\pi]$

As $x\to\left(\tfrac{3\pi}{2}\right)^+$, $\tan x\to -\infty$, so $f(x)\to -\infty.$ At $x=2\pi$, $f(2\pi)=2(2\pi)+3\tan(2\pi)-\pi=4\pi+0-\pi=3\pi>0.$ Hence exactly one solution in this interval.

5. Total number of solutions: $1+1+1+1+1=5.$

6. Therefore the correct option is $\boxed{\text{B: }5}.$