1.
\(4sinx-3cosx=2\)
\(\Leftrightarrow\dfrac{4}{5}sinx-\dfrac{3}{5}cosx=\dfrac{2}{5}\)
Đặt \(\alpha=arccos\left(\dfrac{4}{5}\right)\), phương trình đã cho tương đương:
\(\Leftrightarrow sin\left(x-\alpha\right)=\dfrac{2}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\alpha=arcsin\left(\dfrac{2}{5}\right)+k2\pi\\x-\alpha=\pi-arcsin\left(\dfrac{2}{5}\right)+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-arccos\left(\dfrac{4}{5}\right)=arcsin\left(\dfrac{2}{5}\right)+k2\pi\\x-arccos\left(\dfrac{4}{5}\right)=\pi-arcsin\left(\dfrac{2}{5}\right)+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=arccos\left(\dfrac{4}{5}\right)+arcsin\left(\dfrac{2}{5}\right)+k2\pi\\x=\pi+arccos\left(\dfrac{4}{5}\right)-arcsin\left(\dfrac{2}{5}\right)+k2\pi\end{matrix}\right.\)
3.
\(\sqrt{3}sin3x+cos3x=1\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sin3x+\dfrac{1}{2}cos3x=\dfrac{1}{2}\)
\(\Leftrightarrow sin\left(3x+\dfrac{\pi}{6}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{\pi}{6}=\dfrac{\pi}{6}+k2\pi\\3x+\dfrac{\pi}{6}=\pi-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k2\pi}{3}\\x=\dfrac{2\pi}{9}+\dfrac{k2\pi}{3}\end{matrix}\right.\)
3.
\(5cos2x-12cos2x=13\)
\(\Leftrightarrow-7cos2x=13\)
\(\Leftrightarrow cos2x=-\dfrac{13}{7}\)
\(\Rightarrow\) phương trình vô nghiệm