Number of solutions of equation ${\tan }^{3}x+{\cot }^{3}x+8={(\tan x+\cot x+2)}^3$  in  $[0,4\pi ]$ is ______

 $$\begin{gathered} a^3+b^3+c^3={(a+b+c)}^3=a^3+b^3+c^3+3(a+b)(b+c)(c+c) \\ \Rightarrow (a+b)(b+c)(c+a)=0 \end{gathered}$$
(tan x + cot x) (tan x + 2) (cot x + 2) = 0
tan x + cot x = 0 (or) tan x + 2 = 0 (or) cot x + 2 = 0
${\tan }^{2}x=-1$  (or)  $\tan x=-2$  (or)  $\cot x=-2$
No solution         4 solutions             4 solutions