Đáp án:

92) $\left[\begin{array}{l}x = k\dfrac{\pi}{3}\\x = \pm \dfrac{\pi}{4} + k\dfrac{2\pi}{3}\end{array}\right.\qquad (k\in \Bbb Z)$

93) $x = \pm \dfrac{\pi}{6} + k\pi \qquad \qquad (k \in \Bbb Z)$

94) $\left[\begin{array}{l}x = \dfrac{\pi}{6} + k2\pi\\x = \dfrac{5\pi}{6} + k2\pi\end{array}\right.\qquad \quad(k \in \Bbb Z)$

Giải thích các bước giải:

$\begin{array}{l}92) \, 2\sin3x + \sqrt2\sin6x = 0\\ \Leftrightarrow 2\sin3x + 2\sqrt2\sin3x\cos3x = 0\\ \Leftrightarrow \sin3x(1 + \sqrt2\cos3x) = 0\\ \Leftrightarrow \left[\begin{array}{l}\sin3x = 0\\\cos3x = -\dfrac{\sqrt2}{2}\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}3x = k\pi\\3x = \pm \dfrac{3\pi}{4} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = k\dfrac{\pi}{3}\\x = \pm \dfrac{\pi}{4} + k\dfrac{2\pi}{3}\end{array}\right.\qquad (k\in \Bbb Z)\\ 93)\, 3\cos^2x - 5\sin^2x = 1\\ \Leftrightarrow 3(1 - \sin^2x) - 5\sin^2x = 1\\ \Leftrightarrow -4\sin^2x +1 = 0\\ \Leftrightarrow -2(1 - \cos2x) + 1 = 0\\ \Leftrightarrow 2\cos2x - 1 =0\\ \Leftrightarrow \cos2x = \dfrac{1}{2}\\ \Leftrightarrow 2x = \pm \dfrac{\pi}{3} + k2\pi\\ \Leftrightarrow x = \pm \dfrac{\pi}{6} + k\pi \qquad (k \in \Bbb Z)\\ 94)\,\sin\left(x - \dfrac{\pi}{3}\right) + \sqrt3\cos\left(x - \dfrac{\pi}{3}\right) =1\\ \Leftrightarrow \dfrac{1}{2}\sin\left(x - \dfrac{\pi}{3}\right) +\dfrac{\sqrt3}{2}\cos\left(x - \dfrac{\pi}{3}\right) =\dfrac{1}{2}\\ \Leftrightarrow \cos\dfrac{\pi}{3}\sin\left(x - \dfrac{\pi}{3}\right) +\sin\dfrac{\pi}{3}\cos\left(x - \dfrac{\pi}{3}\right) =\sin\dfrac{\pi}{6}\\ \Leftrightarrow \sin\left(x - \dfrac{\pi}{3} + \dfrac{\pi}{3}\right)= \sin\dfrac{\pi}{6}\\ \Leftrightarrow \sin x = \sin\dfrac{\pi}{6}\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{6} + k2\pi\\x = \dfrac{5\pi}{6} + k2\pi\end{array}\right.\qquad (k \in \Bbb Z) \end{array}$