Ta có: \({\sin}^2 x+{\sin}^2 2x={\sin}^2 3x\)
\(\Leftrightarrow \dfrac{1-\cos 2x}{2}+\dfrac{1-\cos 4x}{2}=\dfrac{1-\cos 6x}{2}\)
\(\Leftrightarrow 1-\cos 4x+\cos 6x-\cos 2x=0\)
\(\Leftrightarrow 2{\sin}^2 2x-2\sin 4x\sin 2x=0\)
\(\Leftrightarrow 2\sin 2x(\sin 2x-\sin 4x)=0\)
\(\Leftrightarrow 2\sin 2x(-2)\cos 3x\sin x=0\)
\(\Leftrightarrow \sin 2x\cos 3x\sin x=0\)
\(\Leftrightarrow \left[ \begin{array}{l} \sin 2x = 0\\\cos 3x= 0\\\sin x=0\end{array} \right. \)
\(\Leftrightarrow \left[ \begin{array}{l} \sin 2x = 0\\\cos 3x= 0\end{array} \right. \)
\(\Leftrightarrow \left[ \begin{array}{l} 2x = k\pi,k\in\mathbb{Z}\\3x=\dfrac{\pi}{2}+k\pi,k\in\mathbb{Z}\end{array} \right. \)
\(\Leftrightarrow \left[ \begin{array}{l} x = k\dfrac{\pi}{2},k\in\mathbb{Z}\\x=\dfrac{\pi}{6}+k\dfrac{\pi}{3},k\in\mathbb{Z}\end{array} \right. \)
Vậy nghiệm của phương là \(x = k\dfrac{\pi}{2},k\in\mathbb{Z}\) và \(x=\dfrac{\pi}{6}+k\dfrac{\pi}{3},k\in\mathbb{Z}\).
