Above question can be reduced to

Find number of points of intersection of graphs of tan x and 2x

Here is the graph of tan x and 2x superimposed

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Now, let's prove there will be 3 intersections on the range $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$

Now, it is evident that at $x = 0, tan x = 2x = 0$ , so there will be an intersection at $x = 0$

After that, $\text{let } f(x) = tan x \text{ and } g(x) = 2x$

$f'(x) = sec²x \text{ and } g'(x) = 2$

$\text{So, at } x = 0, f'(x) \lt g'(x)$

$\text{But, for } x \geq \frac{\pi}{2}, f'(x) \geq g'(x)$

So, there will be another intersection for $x \in (0, \frac{\pi}{2}]$

Similarly, it can be proved that there will be another intersection for $x \in [-\frac{\pi}{2}, 0)$

So, there are three intersections for $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$

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Now, for all other $x \in [\frac{(2k - 1)\pi}{2}, \frac{(2k + 1)\pi}{2}]$ where $k \in I - [0]$ , there is 1 intersection

So, total intersections are $n = 2(15) + 3$

$\color{#3D99F6}{\boxed{n = 33}}$