$\begin{array}{l}
2\cos x + \dfrac{1}{3}{\cos ^2}\left( {x + 3\pi } \right) = \dfrac{8}{3} + \sin \left( {2x - 2\pi } \right) + 3\cos \left( {x + \dfrac{{21\pi }}{2}} \right) + \dfrac{1}{3}{\sin ^2}x\\
 \Leftrightarrow 2\cos x + \dfrac{1}{3}{\cos ^2}\left( {x + \pi } \right) = \dfrac{8}{3} + \sin 2x + 3\cos \left( {x + \dfrac{\pi }{2}} \right) + \dfrac{1}{3}{\sin ^2}x\\
 \Leftrightarrow 2\cos x + \dfrac{1}{3}{\cos ^2}x = \dfrac{8}{3} + \sin 2x - 3\sin x + \dfrac{1}{3}{\sin ^2}x\\
 \Leftrightarrow 6\cos x + {\cos ^2}x - {\sin ^2}x - 3\sin 2x + 9\sin x = 8\\
 \Leftrightarrow 6\cos x + \cos 2x - 3\sin 2x + 9\sin x = 8\\
 \Leftrightarrow 9\sin x + 6\cos x - 6\sin x\cos x + 1 - 2{\sin ^2}x - 8 = 0\\
 \Leftrightarrow 6\cos x\left( {1 - \sin x} \right) + \left( {1 - \sin x} \right)\left( {2\sin x - 7} \right) = 0\\
 \Leftrightarrow \left( {1 - \sin x} \right)\left( {6\cos x + 2\sin x - 7} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin x = 1\\
6\cos x + 2\sin x = 7\left( 1 \right)
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k2\pi \\
6\cos x + 2\sin x = 7
\end{array} \right.\\
\left( 1 \right):a = 6,b = 2,c = 7 \Rightarrow {a^2} + {b^2} = 40 < {c^2} = 49\\
 \Rightarrow PTVN\\
2)DKXD:\cos x \ne 0 \Leftrightarrow x \ne \dfrac{\pi }{2} + k\pi \\
 + \sin x = 1 \Rightarrow \cos x = 0 \Rightarrow VL\\
 + \sin x \ne 1 \Leftrightarrow x \ne \dfrac{\pi }{2} + k2\pi \\
PT \Leftrightarrow {\tan ^2}x - {\tan ^2}x{\sin ^3}x + {\cos ^3}x - 1 = 0\\
 \Leftrightarrow {\tan ^2}x\left( {1 - {{\sin }^3}x} \right) = 1 - {\cos ^3}x\\
 \Leftrightarrow {\tan ^2}x = \dfrac{{1 - {{\cos }^3}x}}{{1 - {{\sin }^3}x}}\\
 \Leftrightarrow \dfrac{{1 - {{\cos }^2}x}}{{1 - {{\sin }^2}x}} = \dfrac{{1 - {{\cos }^3}x}}{{1 - {{\sin }^3}x}}\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x = 1\\
\dfrac{{1 + \cos x}}{{1 + \sin x}} = \dfrac{{1 + {{\cos }^2}x + \cos x}}{{1 + \sin x + {{\sin }^2}x}}\left( {\cos x \ne 1} \right)\left( * \right)
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = k2\pi \\
\dfrac{{1 + \cos x}}{{1 + \sin x}} = \dfrac{{1 + {{\cos }^2}x + \cos x}}{{1 + \sin x + {{\sin }^2}x}}
\end{array} \right.\\
\left( * \right) \Leftrightarrow \left( {1 + \cos x} \right)\left( {1 + \sin x + {{\sin }^2}x} \right)\left( {1 + \sin x} \right)\left( {1 + \cos x + {{\cos }^2}x} \right)\\
 \Leftrightarrow {\sin ^2}x + \cos x{\sin ^2}x = {\cos ^2}x + \sin x{\cos ^2}x\\
 \Leftrightarrow {\sin ^2}x - {\cos ^2}x + \cos x{\sin ^2}x - \sin {\cos ^2}x = 0\\
 \Leftrightarrow \left( {\sin x - \cos x} \right)\left( {\sin x + \cos x} \right) + \cos x\sin x\left( {\sin x - \cos x} \right) = 0\\
 \Leftrightarrow \left( {\sin x - \cos x} \right)\left( {\sin x + \cos x + \sin x\cos x} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin x - \cos x = 0\\
\sin x + \cos x + \sin x\cos x = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt 2 \sin \left( {x - \dfrac{\pi }{4}} \right) = 0\\
t + \dfrac{{{t^2} - 1}}{2} = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{\pi }{4} = k\pi \\
{t^2} + 2t - 1 = 0\left( { - \sqrt 2  \le t = \sin x + \cos x \le \sqrt 2 } \right)
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
t = \sqrt 2  - 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \left( {k \ne 2m,m \in \mathbb{Z}} \right)\\
t = \sqrt 2  - 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
\sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) = \sqrt 2  - 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
\sin \left( {x + \dfrac{\pi }{4}} \right) = 1 - \dfrac{{\sqrt 2 }}{2}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x + \dfrac{\pi }{4} = \arcsin \left( {1 - \dfrac{{\sqrt 2 }}{2}} \right) + k2\pi \\
x + \dfrac{\pi }{4} = \pi  - \arcsin \left( {1 - \dfrac{{\sqrt 2 }}{2}} \right) + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x =  - \dfrac{\pi }{4} + \arcsin \left( {1 - \dfrac{{\sqrt 2 }}{2}} \right) + k2\pi \\
x = \dfrac{{3\pi }}{4} + \arcsin \left( {1 - \dfrac{{\sqrt 2 }}{2}} \right) + k2\pi 
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = k2\pi \\
x = \dfrac{\pi }{4} + k\pi \\
x =  - \dfrac{\pi }{4} + \arcsin \left( {1 - \dfrac{{\sqrt 2 }}{2}} \right) + k2\pi \\
x = \dfrac{{3\pi }}{4} + \arcsin \left( {1 - \dfrac{{\sqrt 2 }}{2}} \right) + k2\pi 
\end{array} \right.\left( {k \in \mathbb{Z}} \right)
\end{array}$