a) ta có : \(2sin^2x+3cos2x=0\Leftrightarrow2sin^2x+3\left(1-2sin^2x\right)=0\)
\(\Leftrightarrow3-4sin^2x=0\Leftrightarrow sin^2x=\dfrac{3}{4}\Leftrightarrow sinx=\pm\dfrac{\sqrt{3}}{2}\)
th1 : \(sinx=\dfrac{\sqrt{3}}{2}\Leftrightarrow sinx=sin\dfrac{\pi}{3}\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k2\pi\\x=\pi-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k2\pi\\x=\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)th2 : \(sinx=\dfrac{-\sqrt{3}}{2}\Leftrightarrow sinx=sin\left(\dfrac{-\pi}{3}\right)\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\pi}{3}+k2\pi\\x=\pi+\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\pi}{3}+k2\pi\\x=\dfrac{4\pi}{3}+k2\pi\end{matrix}\right.\)
vậy phương trình có 4 hệ nghiệm : \(x=\dfrac{\pi}{3}+k2\pi;x=\dfrac{2\pi}{3}+k2\pi;x=\dfrac{-\pi}{3}+k2\pi;x=\dfrac{4\pi}{3}+k2\pi\)
câu b bn làm tương tự cho quen nha
