\(3\sin x-4\cos x=1\Leftrightarrow\dfrac{3}{5}\sin x-\dfrac{4}{5}\cos x=\dfrac{1}{5}\)
\(\Leftrightarrow\sin\left(x-\alpha\right)=\dfrac{1}{5}\) (với \(\cos\alpha=\dfrac{3}{5};\sin\alpha=\dfrac{4}{5}\) )
\(\Leftrightarrow\left[{}\begin{matrix}x=\alpha+arc\sin\dfrac{1}{5}+k2\pi,k\in\mathbb{Z}\\x=\alpha+\pi-arc\sin\dfrac{1}{5}+k2\pi,k\in\mathbb{Z}\end{matrix}\right.\)