Đáp án:
\[\left[ \begin{array}{l}
x = k\pi \\
x = \pm \dfrac{\pi }{3} + k\pi
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\cos 2x = 1 - 2{\sin ^2}x \Rightarrow {\sin ^2}x = \dfrac{{1 - \cos 2x}}{2}\\
{\sin ^2}\left( {x - \dfrac{{2\pi }}{3}} \right) + {\sin ^2}x + {\sin ^2}\left( {x + \dfrac{{2\pi }}{3}} \right) = \dfrac{3}{2}\\
\Leftrightarrow \dfrac{{1 - \cos \left( {2x - \dfrac{{4\pi }}{3}} \right)}}{2} + \dfrac{{1 - \cos 2x}}{2} + \dfrac{{1 - \cos \left( {2x + \dfrac{{4\pi }}{3}} \right)}}{2} = \dfrac{3}{2}\\
\Leftrightarrow \dfrac{3}{2} - \dfrac{1}{2}\left( {\cos \left( {2x - \dfrac{{4\pi }}{3}} \right) + \cos 2x + \cos \left( {2x + \dfrac{{4\pi }}{3}} \right)} \right) = \dfrac{3}{2}\\
\Leftrightarrow \cos \left( {2x - \dfrac{{4\pi }}{3}} \right) + \cos 2x + \cos \left( {2x + \dfrac{{4\pi }}{3}} \right) = 0\\
\Leftrightarrow \left[ {\cos \left( {2x - \dfrac{{4\pi }}{3}} \right) + \cos \left( {2x + \dfrac{{4\pi }}{3}} \right)} \right] + \cos 2x = 0\\
\Leftrightarrow 2.\cos \left( {2x - \dfrac{{4\pi }}{3} + 2x + \dfrac{{4\pi }}{3}} \right).\cos \left( {2x - \dfrac{{4\pi }}{3} - 2x - \dfrac{{4\pi }}{3}} \right) + \cos 2x = 0\\
\Leftrightarrow 2.\cos 4x.\cos \left( { - \dfrac{{8\pi }}{3}} \right) + \cos 2x = 0\\
\Leftrightarrow 2.\cos 4x.\left( { - \dfrac{1}{2}} \right) + \cos 2x = 0\\
\Leftrightarrow \cos 4x - \cos 2x = 0\\
\Leftrightarrow \left( {2{{\cos }^2}2x - 1} \right) - \cos 2x = 0\\
\Leftrightarrow \left( {\cos 2x - 1} \right)\left( {2\cos 2x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 2x = 1\\
\cos 2x = - \dfrac{1}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x = k2\pi \\
2x = \pm \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \pm \dfrac{\pi }{3} + k\pi
\end{array} \right.
\end{array}\)
