Đáp án:
$3a)\, \left[\begin{array}{l}x = \dfrac{\pi}{3} + k2\pi\\x = \dfrac{2\pi}{9} = k\dfrac{2\pi}{3}\end{array}\right.\,\,\,\,\,(k \in \Bbb Z)$
$3b)\, \left[\begin{array}{l}x = \dfrac{\arccos\dfrac{2}{\sqrt{13}}}{6} + k\dfrac{\pi}{3}\\x = \dfrac{\pi}{8} - \dfrac{\arccos\dfrac{2}{\sqrt{13}}}{8} + k\dfrac{\pi}{4}\end{array}\right.\,\,\,\,\,(k \in \Bbb Z)$
Giải thích các bước giải:
$\begin{array}{l}3a)\, \sin3x + \sqrt3\cos3x = 2\sin2x\\ \Leftrightarrow \dfrac{1}{2}\sin3x + \dfrac{\sqrt3}{2}\cos3x = \sin2x\\ \Leftrightarrow \cos\dfrac{\pi}{3}\sin3x + \sin\dfrac{\pi}{3}\cos3x = \sin2x\\ \Leftrightarrow \sin\left(x + \dfrac{\pi}{3}\right) = \sin2x\\ \leftrightarrow \left[\begin{array}{l}x + \dfrac{\pi}{3} = 2x + k2\pi\\x + \dfrac{\pi}{3} = \pi - 2x + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{3} + k2\pi\\x = \dfrac{2\pi}{9} = k\dfrac{2\pi}{3}\end{array}\right.&(k \in \Bbb Z)\\ 3b)\, 3\cos x + 2\sin x = \sqrt{13}\sin 7x\\ \Leftrightarrow \dfrac{3}{\sqrt{13}}\cos x + \dfrac{2}{\sqrt{13}}\sin x = \sin 7x&(*)\\ Đặt:\quad \cos\alpha = \dfrac{2}{\sqrt{13}}\Rightarrow \alpha = \arccos\dfrac{2}{\sqrt{13}}\\ Do\quad \sin^2\alpha + \cos^2\alpha = 1, \forall \alpha\\ \Rightarrow Đặt\quad \sin\alpha = \dfrac{3}{\sqrt{13}}\\ (*) \Leftrightarrow \sin\alpha\cos x + \cos\alpha\sin x = \sin7x\\ \Leftrightarrow \sin(x + \alpha) = \sin7x\\ \Leftrightarrow \left[\begin{array}{l}x + \alpha = 7x + k2\pi\\x + \alpha = \pi - 7x + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\alpha}{6} + k\dfrac{\pi}{3}\\x = \dfrac{\pi}{8} - \dfrac{\alpha}{8} + k\dfrac{\pi}{4}\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\arccos\dfrac{2}{\sqrt{13}}}{6} + k\dfrac{\pi}{3}\\x = \dfrac{\pi}{8} - \dfrac{\arccos\dfrac{2}{\sqrt{13}}}{8} + k\dfrac{\pi}{4}\end{array}\right.&(k \in \Bbb Z) \end{array}$
