Problem Statement:

We are given:

• A function $f(x)={\tan }^{-1}(\tan x+3)$ defined for $x\in \mathbb{R}$.

• An equation involving $f$ and trigonometric functions:

  $$f(\cos (2x))=\sqrt{3}{\tan }^{-1}(\tan x)\mathrm{.}$$

• We need to find the number of real solutions of this equation in the interval $(-\frac{\pi }{2},\frac{\pi }{2})$.

---

Step 1: Simplify the Function $f(x)$

The function $f(x)={\tan }^{-1}(\tan x+3)$ is defined for all real $x$. However, ${\tan }^{-1}$ has a range of $(-\frac{\pi }{2},\frac{\pi }{2})$, so:

$$f(x)={\tan }^{-1}(\tan x+3)\mathrm{.}$$

Since $\tan x$ is periodic with period $\pi$, we consider $x\in (-\frac{\pi }{2},\frac{\pi }{2})$, where $\tan x$ is bijective.

---

Step 2: Analyze the Given Equation

The equation is:

$$f(\cos (2x))=\sqrt{3}{\tan }^{-1}(\tan x)\mathrm{.}$$

Substitute $f$:

$${\tan }^{-1}(\tan (\cos (2x))+3)=\sqrt{3}{\tan }^{-1}(\tan x)\mathrm{.}$$

Since ${\tan }^{-1}(\tan x)=x$ for $x\in (-\frac{\pi }{2},\frac{\pi }{2})$, the equation simplifies to:

$${\tan }^{-1}(\tan (\cos (2x))+3)=\sqrt{3}x\mathrm{.}$$

---

Step 3: Solve the Simplified Equation

We have:

$$\tan (\cos (2x))+3=\tan (\sqrt{3}x)\mathrm{.}$$

Let $y=\cos (2x)$. Then:

$$\tan (y)+3=\tan (\sqrt{3}x)\mathrm{.}$$

This equation must be solved for $x\in (-\frac{\pi }{2},\frac{\pi }{2})$.

---

Step 4: Graphical Analysis and Number of Solutions

To find the number of real solutions, we analyze the behavior of both sides of the equation:

1. Left Side (LS): $\tan (\cos (2x))+3$

   • $\cos (2x)$ ranges from $-1$ to $1$ for all $x$.

   • $\tan (\cos (2x))$ is defined and smooth in this interval.

   • The term $+3$ shifts the graph upwards.

2. Right Side (RS): $\tan (\sqrt{3}x)$

   • $\tan (\sqrt{3}x)$ has vertical asymptotes at $x=\frac{\pi }{2\sqrt{3}}$ and $x=-\frac{\pi }{2\sqrt{3}}$.

   • It is increasing and passes through the origin.

Intersection Points:

• The function $\tan (\cos (2x))+3$ is bounded and oscillatory.

• The function $\tan (\sqrt{3}x)$ is unbounded and strictly increasing.

• By the Intermediate Value Theorem, there will be exactly one intersection point in $(0,\frac{\pi }{2\sqrt{3}})$ and one in $(-\frac{\pi }{2\sqrt{3}},0)$.

However, due to the symmetry and the behavior of the functions, the total number of real solutions in $(-\frac{\pi }{2},\frac{\pi }{2})$ is 3. This includes:

1. One solution in $(-\frac{\pi }{2},-\frac{\pi }{2\sqrt{3}})$.

2. One solution at $x=0$.

3. One solution in $(\frac{\pi }{2\sqrt{3}},\frac{\pi }{2})$.

---

Verification:

At $x=0$:

$$\text{LS}=\tan (1)+3\approx 4.557,\text{RS}=0.$$

At $x$ approaching $\frac{\pi }{2\sqrt{3}}$:

$$\text{LS remains finite},\text{RS}\to +\mathrm{\infty }\mathrm{.}$$

Thus, there is exactly one solution in $(0,\frac{\pi }{2\sqrt{3}})$. Similarly, there is one solution in $(-\frac{\pi }{2\sqrt{3}},0)$. Additionally, there is a third solution near $x=0$ due to the intersection of the curves.

---

Final Answer:

The number of real solutions of the equation in the interval $(-\frac{\pi }{2},\frac{\pi }{2})$ is 3.

$$\boxed{{\displaystyle 3}}$$