Giải phương trình \(\cos x-\sqrt{3}\sin x=\sqrt{2}\).
\(\left[\begin{array}{nghiempt}x=-\frac{\pi}{12}+2k\pi\\x=-\frac{7\pi}{12}+2k\pi\end{array}\right.\) (\(k\in Z\)).\(\left[\begin{array}{nghiempt}x=\frac{\pi}{4}+2k\pi\\x=\frac{3\pi}{4}+2k\pi\end{array}\right.\) (\(k\in Z\)).\(\left[\begin{array}{nghiempt}x=\frac{\pi}{12}+2k\pi\\x=\frac{7\pi}{12}+2k\pi\end{array}\right.\) (\(k\in Z\)).\(\left[\begin{array}{nghiempt}x=-\frac{\pi}{4}+2k\pi\\x=-\frac{3\pi}{4}+2k\pi\end{array}\right.\) (\(k\in Z\)).Hướng dẫn giải:Chia hai vế cho \(\sqrt{1^2+\sqrt{3}^2}=2\) ta được:
\(\frac{1}{2}\cos x-\frac{\sqrt{3}}{2}\sin x=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\sin\frac{\pi}{6}\cos x-\cos\frac{\pi}{6}\sin x=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\sin\left(\frac{\pi}{6}-x\right)=\sin\frac{\pi}{4}\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\frac{\pi}{6}-x=\frac{\pi}{4}+2k\pi\\\frac{\pi}{6}-x=\pi-\frac{\pi}{4}+2k\pi\end{array}\right.\) (với \(k\in\mathbb{Z}\))
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{\pi}{12}-2k\pi\\x=-\frac{7\pi}{12}-2k\pi\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{\pi}{12}+2k\pi\\x=-\frac{7\pi}{12}+2k\pi\end{array}\right.\) (thay k bởi -k thì tập nghiệm không thay đổi)