Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\cos x + \cos y = 2\cos \frac{{x + y}}{2}.\cos \frac{{x - y}}{2}\\
2\cos x.\cos y = \cos \left( {x + y} \right) + \cos \left( {x - y} \right)\\
{\cos ^2}x + {\cos ^2}\left( {x + \frac{{2\pi }}{3}} \right) + {\cos ^2}\left( {x + \frac{{4\pi }}{3}} \right)\\
\cos 2x = 2{\cos ^2}x - 1\\
= \left[ {{{\cos }^2}x + {{\cos }^2}\left( {x + \frac{{4\pi }}{3}} \right)} \right] + {\cos ^2}\left( {x + \frac{{2\pi }}{3}} \right)\\
= {\left[ {\cos x + \cos \left( {x + \frac{{4\pi }}{3}} \right)} \right]^2} - 2.\cos x.\cos \left( {x + \frac{{4\pi }}{3}} \right) + {\cos ^2}\left( {x + \frac{{2\pi }}{3}} \right)\\
= {\left[ {2.\cos \left( {x + \frac{{2\pi }}{3}} \right).cos\frac{{2\pi }}{3}} \right]^2} - \left[ {\cos \left( {2x + \frac{{4\pi }}{3}} \right) + \cos \frac{{4\pi }}{3}} \right] + {\cos ^2}\left( {x + \frac{{2\pi }}{3}} \right)\\
= 4{\cos ^2}\left( {x + \frac{{2\pi }}{3}} \right).{\cos ^2}\frac{{2\pi }}{3} - \cos \left( {2x + \frac{{4\pi }}{3}} \right) - \cos \frac{{4\pi }}{3} + {\cos ^2}\left( {x + \frac{{2\pi }}{3}} \right)\\
= {\cos ^2}\left( {x + \frac{{2\pi }}{3}} \right) - \left( {2{{\cos }^2}\left( {x + \frac{{2\pi }}{3}} \right) - 1} \right) + \frac{1}{2} + {\cos ^2}\left( {x + \frac{{2\pi }}{3}} \right)\\
= \frac{3}{2}
\end{array}\)
