Giải phương trình sau
\({\sin ^2}x + {\sin ^2}2x = {\sin ^2}3x\)
Ta có: \({\sin ^2}x + {\sin ^2}2x = {\sin ^2}3x\)
\(\begin{array}{l}
\Leftrightarrow \frac{{1 - \cos 2x}}{2} + \frac{{1 - \cos 4x}}{2} = \frac{{1 - \cos 6x}}{2}\\
\Leftrightarrow 1 - \cos 4x + \cos 6x - \cos 2x = 0\\
\Leftrightarrow 2{\sin ^2}2x - 2\sin 4x\sin 2x = 0\\
\Leftrightarrow 2\sin 2x(\sin 2x - \sin 4x) = 0\\
\Leftrightarrow 2\sin 2x( - 2)\cos 3x\sin x = 0\\
\Leftrightarrow \sin 2x\cos 3x\sin x = 0\\
\Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{\sin 2x = 0}\\
{\cos 3x = 0}\\
{\sin x = 0}
\end{array}} \right.\\
\Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{\sin 2x = 0}\\
{\cos 3x = 0}
\end{array}} \right.\\
\Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{2x = k\pi ,k \in Z}\\
{3x = \frac{\pi }{2} + k\pi ,k \in Z}
\end{array}} \right.\\
\Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x = k\frac{\pi }{2},k \in Z}\\
{x = \frac{\pi }{6} + k\frac{\pi }{3},k \in Z}
\end{array}} \right.
\end{array}\)
-- Mod Toán 11