显然 $x\ne 0$，考虑函数\[g(x)=\dfrac{\ln\tan x+\cos 2x}x,\]则其导函数\[g'(x)=\dfrac{-\cos 2x+\dfrac{2x\cos^22x}{\sin 2x}-\ln\tan x}{x^2},\]设分子部分为 $\varphi(x)$，则注意到\[\varphi\left(\dfrac{\pi}4\right)=0,\]于是考虑其导函数\[\varphi'(x)=\dfrac{4x\cos 2x(\cos^22x-2)}{\sin^22x},\]于是\[\begin{array} {c|ccc}\hline
x&\left(0,\dfrac{\pi}4\right)&\dfrac{\pi}4&\left(\dfrac{\pi}4,\dfrac{\pi}2\right)\\ \hline
\varphi'(x)&-&0&+\\ \hline
\varphi(x)&+&0&+\\ \hline\end{array}\]因此函数 $g(x)$ 在 $\left(0,\dfrac{\pi}2\right)$ 单调递增．类似的，可得函数 $g(x)$ 在 $\left[-5,-\dfrac{3\pi}2\right)$，$\left(-\pi,-\dfrac{\pi}2\right)$ 均单调递减，在 $\left(\pi,\dfrac{3\pi}2\right)$ 上单调递增．注意到\[\begin{array} {c|cccccccccccc}\hline
x&-\dfrac{7\pi}4&-5&-\dfrac{3\pi}2&-\pi&-\dfrac{3\pi}4&-\dfrac{\pi}2&0&\dfrac{\pi}4&\dfrac{\pi}2&\pi&\dfrac{5\pi}4&\dfrac{3\pi}2 \\ \hline
g(x)&0&-&-\infty&+\infty&0&-\infty&-\infty&0&+\infty&-\infty&0&+\infty\\ \hline
\end{array}\]于是方程 $g(x)=2$ 在 $[-5,5]$ 上有 $3$ 个公共点．