Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\cos 2x = 1 - 2{\sin ^2}x \Rightarrow {\sin ^2}x = \dfrac{{1 - \cos 2x}}{2}\\
{\sin ^2}x + {\sin ^2}\left( {x + \dfrac{\pi }{3}} \right) + {\sin ^2}\left( {x - \dfrac{\pi }{3}} \right)\\
= \dfrac{{1 - \cos 2x}}{2} + \dfrac{{1 - \cos \left( {2x + \dfrac{{2\pi }}{3}} \right)}}{2} + \dfrac{{1 - \cos \left( {2x - \dfrac{{2\pi }}{3}} \right)}}{2}\\
= \dfrac{3}{2} - \dfrac{1}{2}.\left[ {\cos 2x + \cos \left( {2x + \dfrac{{2\pi }}{3}} \right) + \cos \left( {2x - \dfrac{{2\pi }}{3}} \right)} \right]\\
= \dfrac{3}{2} - \dfrac{1}{2}.\left[ {\cos 2x + 2.\cos \dfrac{{2x + \dfrac{{2\pi }}{3} + 2x - \dfrac{{2\pi }}{3}}}{2}.\cos \dfrac{{2x + \dfrac{{2\pi }}{3} - 2x + \dfrac{{2\pi }}{3}}}{2}} \right]\\
= \dfrac{3}{2} - \dfrac{1}{2}.\left[ {\cos 2x + 2.\cos 2x.\cos \dfrac{{2\pi }}{3}} \right]\\
= \dfrac{3}{2} - \dfrac{1}{2}\left[ {\cos 2x + 2.\cos 2x.\left( { - \dfrac{1}{2}} \right)} \right]\\
= \dfrac{3}{2} - \dfrac{1}{2}.0 = \dfrac{3}{2}
\end{array}\)
