Đáp án:
$\begin{array}{l}a)\,\,\left[\begin{array}{l}x= \pi - 4\arccos\dfrac{\sqrt2}{\sqrt3} + k4\pi\\x = \pi + k4\pi\end{array}\right.\quad (k \in \Bbb Z)\\b)\,\,\left[\begin{array}{l}x = 2\arccos\dfrac{2}{\sqrt{29}} + k2\pi\\x = \pi + k2\pi\end{array}\right.\quad (k \in \Bbb Z)\\c)\,\,x = \dfrac{\pi}{6} + \dfrac{1}{3}\arccos\dfrac{3}{5} + k\dfrac{2\pi}{3}\quad (k\in\Bbb Z)\end{array}$
Giải thích các bước giải:
$\begin{array}{l}a)\,\,\sqrt2\sin\dfrac{x}{2} + \cos\dfrac{x}{2} = \sqrt2\\ \Leftrightarrow \dfrac{\sqrt2}{\sqrt3}\sin\dfrac{x}{2} + \dfrac{1}{\sqrt3}\cos\dfrac{x}{2} = \dfrac{\sqrt2}{\sqrt3}\\ Do \,\,\left(\dfrac{\sqrt2}{\sqrt3}\right)^2 + \left(\dfrac{1}{\sqrt3}\right)^2 = 1\\ Đặt\,\,\begin{cases}\cos\alpha = \dfrac{\sqrt2}{\sqrt3}\\\sin\alpha = \dfrac{1}{\sqrt3}\end{cases}\Rightarrow \alpha = \arccos\dfrac{\sqrt2}{\sqrt3}\\ Ta\,\,được:\\ \sin\dfrac{x}{2}\cos\alpha + \cos\dfrac{x}{2}\sin\alpha = \cos\alpha\\ \Leftrightarrow \sin\left(\dfrac{x}{2} + \alpha\right) = \sin\left(\dfrac{\pi}{2} - \alpha\right)\\ \Leftrightarrow \left[\begin{array}{l}\dfrac{x}{2} + \alpha = \dfrac{\pi}{2} - \alpha + k2\pi\\\dfrac{x}{2} + \alpha = \dfrac{\pi}{2} + \alpha + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}\dfrac{x}{2} = \dfrac{\pi}{2} - 2\alpha + k2\pi\\\dfrac{x}{2} = \dfrac{\pi}{2} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x= \pi - 4\alpha + k4\pi\\x = \pi + k4\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x= \pi - 4\arccos\dfrac{\sqrt2}{\sqrt3} + k4\pi\\x = \pi + k4\pi\end{array}\right.\quad (k \in \Bbb Z)\\ b)\,\,2\sin x - 5\cos x = 5\\ \Leftrightarrow \dfrac{2}{\sqrt{29}}\sin x - \dfrac{5}{\sqrt{29}}\cos x = \dfrac{5}{\sqrt{29}}\\ Do \,\,\left(\dfrac{2}{\sqrt{29}}\right)^2 + \left(\dfrac{5}{\sqrt{29}}\right)^2 = 1\\ Đặt\,\,\begin{cases}\cos\alpha = \dfrac{2}{\sqrt{29}}\\\sin\alpha = \dfrac{5}{\sqrt{29}}\end{cases}\Rightarrow \alpha = \arccos\dfrac{2}{\sqrt{29}}\\ Ta\,\,được:\\ \sin x\cos\alpha - \cos x\sin\alpha = \sin\alpha\\ \Leftrightarrow \sin(x - \alpha) = \sin\alpha\\ \Leftrightarrow \left[\begin{array}{l}x - \alpha = \alpha + k2\pi\\x - \alpha = \pi - \alpha + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = 2\alpha + k2\pi\\x = \pi + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = 2\arccos\dfrac{2}{\sqrt{29}} + k2\pi\\x = \pi + k2\pi\end{array}\right.\quad (k \in \Bbb Z)\\ c) \,\,3\sin3x - 4\cos3x = 5\\ \Leftrightarrow \dfrac{3}{5}\sin3x - \dfrac{4}{5}\cos3x = 1\\ Do \,\,\left(\dfrac{3}{5}\right)^2 + \left(\dfrac{4}{5}\right)^2 = 1\\ Đặt\,\,\begin{cases}\cos\alpha = \dfrac{3}{5}\\\sin\alpha = \dfrac{4}{5}\end{cases}\Rightarrow \alpha = \arccos\dfrac{3}{5}\\ Ta\,\,được:\\ \sin3x\cos\alpha - \cos3x\sin\alpha = 1\\ \Leftrightarrow \sin(3x - \alpha) = 1\\ \Leftrightarrow 3x - \alpha = \dfrac{\pi}{2} +k2\pi\\ \Leftrightarrow 3x = \dfrac{\pi}{2} + \alpha + k2\pi\\ \Leftrightarrow x = \dfrac{\pi}{6} + \dfrac{1}{3}\alpha + k\dfrac{2\pi}{3}\\ \Leftrightarrow x = \dfrac{\pi}{6} + \dfrac{1}{3}\arccos\dfrac{3}{5} + k\dfrac{2\pi}{3}\quad (k\in\Bbb Z)\\ \end{array}$
