a/ Đặt \(sinx=t\Rightarrow-1\le t\le1\)
Phương trình trở thành:
\(2t^2-3t-2=0\Rightarrow\left[{}\begin{matrix}t=2>1\left(l\right)\\t=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow sinx=-\frac{1}{2}=sin\left(-\frac{\pi}{6}\right)\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
b/ \(\Leftrightarrow sinx=-cos3x\)
\(\Leftrightarrow sinx=sin\left(3x-\frac{\pi}{2}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3x-\frac{\pi}{2}+k2\pi\\x=\pi-3x+\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{3\pi}{8}+\frac{k\pi}{2}\end{matrix}\right.\)
