Number of solutions of $$\sqrt{3}\cos2\theta+8\cos\theta+3\sqrt{3}=0,\theta\epsilon[-3\pi,2\pi]$$ is:

$$\sqrt{3}\cos2\theta+8\cos\theta+3\sqrt{3}=0$$

We know that $$\cos2\theta=2cos^2\theta-1$$

$$\sqrt{3}\left(2\cos^2\theta-1\right)+8\cos\theta+3\sqrt{3}=0$$ 

$$2\sqrt{3}\cos^2\theta-\sqrt{3}+8\cos\theta+3\sqrt{3}=0$$

$$2\sqrt{3}\cos^2\theta+8\cos\theta+2\sqrt{3}=0$$

$$2\sqrt{3}\cos^2\theta+6\cos\theta+2\cos\theta+2\sqrt{3}=0$$

$$2\sqrt{3}\cos\theta(\cos\theta+\sqrt{3})+2(\cos\theta+\sqrt{3})=0$$

$$(2\sqrt{3}\cos\theta+2)(\cos\theta+\sqrt{3})=0$$

Thus, there are two solutions possible -> $$\cos\theta=-\dfrac{1}{\sqrt{3}}$$ and $$\cos\theta=-\sqrt{3}$$

Since, the value of $$\cos\theta\epsilon[-1,1]$$, therefore, $$\cos\theta=-\sqrt{3}$$ is never possible. 

Now, $$\cos\theta$$ is negative in the second and the third quadrant, and in each cycle of $$\pi$$, $$\cos\theta$$ takes all values in $$[-1,1]$$. Thus, in every cycle of $$\pi$$, there will be one solution for $$\cos\theta=-\dfrac{1}{\sqrt{3}}$$.

Thus, from $$[-3\pi,2\pi]$$, there will be 5 solutions for this equation.