Đáp án:
\[C = \dfrac{{11}}{2}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\cos 2x = 2{\cos ^2}x - 1 \Rightarrow {\cos ^2}x = \dfrac{{\cos 2x + 1}}{2}\\
\tan x = \dfrac{{\sin x}}{{\cos x}} = \dfrac{{\cos \left( {\dfrac{\pi }{2} - x} \right)}}{{\sin \left( {\dfrac{\pi }{2} - x} \right)}} = \cot \left( {\dfrac{\pi }{2} - x} \right)\\
\cos a + \cos b = 2\cos \dfrac{{a + b}}{2}.\cos \dfrac{{a - b}}{2}\\
C = {\cos ^2}x + {\cos ^2}\left( {\dfrac{{4\pi }}{3} + x} \right) + {\cos ^2}\left( {\dfrac{{4\pi }}{3} - x} \right) - 4\tan \left( {x - \dfrac{\pi }{6}} \right).\tan \left( {x + \dfrac{{7\pi }}{3}} \right)\\
= \dfrac{{\cos 2x + 1}}{2} + \dfrac{{\cos \left( {\dfrac{{8\pi }}{3} + 2x} \right) + 1}}{2} + \dfrac{{\cos \left( {\dfrac{{8\pi }}{3} - 2x} \right) + 1}}{2} - 4\tan \left( {x - \dfrac{\pi }{6}} \right).\tan \left( {x + \dfrac{\pi }{3} + 2\pi } \right)\\
= \dfrac{3}{2} + \dfrac{1}{2}\cos 2x + \dfrac{1}{2}.\left[ {\cos \left( {\dfrac{{8\pi }}{3} + 2x} \right) + \cos \left( {\dfrac{{8\pi }}{3} - 2x} \right)} \right] - 4\tan \left( {x - \dfrac{\pi }{6}} \right).\tan \left( {x + \dfrac{\pi }{3}} \right)\\
= \dfrac{3}{2} + \dfrac{1}{2}.\cos 2x + \dfrac{1}{2}.2.\cos \dfrac{{8\pi }}{3}.cos2x - 4.\tan \left( {x - \dfrac{\pi }{6}} \right).\cot \left( {\dfrac{\pi }{2} - \left( {x + \dfrac{\pi }{3}} \right)} \right)\\
= \dfrac{3}{2} + \dfrac{1}{2}.cos2x + \left( { - \dfrac{1}{2}} \right).\cos 2x - 4.\tan \left( {x - \dfrac{\pi }{6}} \right).\cot \left( {\dfrac{\pi }{6} - x} \right)\\
= \dfrac{3}{2} - 4.\tan \left( {x - \dfrac{\pi }{6}} \right).\left[ { - \cot \left( {x - \dfrac{\pi }{6}} \right)} \right]\\
= \dfrac{3}{2} + 4\\
= \dfrac{{11}}{2}
\end{array}\)
