$3\sin x-4\cos x=1$
$\Rightarrow \dfrac{3}{5}\sin x- \dfrac{4}{5}\cos x=\dfrac{1}{5}$
Đặt $\cos \alpha=\dfrac{3}{5}$ và $\sin \alpha=\dfrac{4}{5}$
$\Rightarrow \cos \alpha\sin x-\sin \alpha\cos x=\dfrac{1}{5}$
$\Rightarrow \sin(x-\alpha)=\dfrac{1}{5}$
$\Rightarrow \left[ \begin{array}{l} x-\alpha=\arcsin\dfrac{1}{5}+k2\pi \\ x-\alpha=\pi- \arcsin\dfrac{1}{5}+k2\pi \end{array} \right .$
$\Rightarrow \left[\begin{array}{l} x=\alpha+\arcsin\dfrac{1}{5}+k2\pi \\ x=\alpha+\pi- \arcsin\dfrac{1}{5}+k2\pi \end{array} \right .(k\in\mathbb Z)$.
