Ta có sin² x – cos² 3x = 0

\( \Leftrightarrow \frac{{1 - co{s^2}x}}{2} - \frac{{1 + cos6{\rm{x}}}}{2} = 0\)

⇔ cos2x + cos6x = 0

⇔ 2cos2x . cos4x = 0

\[ \Leftrightarrow \left[ \begin{array}{l}cos4{\rm{x}} = 0\\cos2{\rm{x}} = 0\end{array} \right.\]

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

 \( \Leftrightarrow \left[ \begin{array}{l}{\rm{x}} = \frac{\pi }{8} + \frac{{k\pi }}{4}\\{\rm{x}} = \frac{\pi }{4} + \frac{{k\pi }}{2}\end{array} \right.\) (k ∈ ℤ)

+) Với \(x = \frac{\pi }{8} + \frac{{k\pi }}{4}\)

Vì 0 ≤ x ≤ π

Nên \(0 \le \frac{\pi }{8} + \frac{{k\pi }}{4} \le \pi \)

\[ \Leftrightarrow - \frac{\pi }{8} \le \frac{{k\pi }}{4} \le \pi - \frac{\pi }{8}\]

⇔ – 0,5 ≤ k ≤ 3,5

Mà k ∈ ℤ nên k ∈ {0; 1; 2; 3}

Khi đó \(x \in \left\{ {\frac{\pi }{8};\frac{{3\pi }}{8};\frac{{5\pi }}{8};\frac{{7\pi }}{8}} \right\}\)

+) Với \(x = \frac{\pi }{4} + \frac{{k\pi }}{2}\)

Vì 0 ≤ x ≤ π

Nên \(0 \le \frac{\pi }{4} + \frac{{k\pi }}{2} \le \pi \)

\( \Leftrightarrow - \frac{\pi }{4} \le \frac{{k\pi }}{2} \le \pi - \frac{\pi }{4}\)

⇔ – 0,5 ≤ k ≤ 1,5

Mà k ∈ ℤ nên k ∈ {0; 1}

Khi đó \(x \in \left\{ {\frac{\pi }{4};\frac{{3\pi }}{4}} \right\}\)

Vậy \(x \in \left\{ {\frac{\pi }{8};\frac{{3\pi }}{8};\frac{{5\pi }}{8};\frac{{7\pi }}{8};\frac{\pi }{4};\frac{{3\pi }}{4}} \right\}\).