Lời giải

Ta có sin²2x + sin²4x = sin²6x.

\( \Leftrightarrow \frac{{1 - \cos 4x}}{2} + \frac{{1 - \cos 8x}}{2} = \frac{{1 - \cos 12x}}{2}\)

⇔ 1 – cos4x + 1 – cos8x = 1 – cos12x

⇔ (cos12x – cos4x) + (1 – cos8x) = 0

⇔ –2sin8x.sin4x + 2sin²4x = 0

⇔ –2sin4x.(sin8x – sin4x) = 0

\( \Leftrightarrow \left[ \begin{array}{l}\sin 4x = 0\\\sin 8x = \sin 4x\end{array} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l}4x = k\pi \\8x = 4x + k2\pi \\8x = \pi - 4x + k2\pi \end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\)

\( \Leftrightarrow \left[ \begin{array}{l}x = k\frac{\pi }{4}\\4x = k2\pi \\12x = \pi + k2\pi \end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\)

\( \Leftrightarrow \left[ \begin{array}{l}x = k\frac{\pi }{4}\\x = k\frac{\pi }{2}\\x = \frac{\pi }{{12}} + k\frac{\pi }{6}\end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\)

\( \Leftrightarrow \left[ \begin{array}{l}x = k\frac{\pi }{4}\\x = \frac{\pi }{{12}} + k\frac{\pi }{6}\end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\).

Vậy phương trình đã cho có nghiệm là \(x = k\frac{\pi }{4};\,\,x = \frac{\pi }{{12}} + k\frac{\pi }{6}\,\,\left( {k \in \mathbb{Z}} \right)\).