Giải phương trình: \(\cos x\cos\dfrac{x}{2}\cos\dfrac{3x}{2}-\sin x\sin\dfrac{x}{2}\sin\dfrac{3x}{2}=\dfrac{1}{2}\).
\(x=\dfrac{\pi}{6}+2k\pi;x=\dfrac{5\pi}{6}+2k\pi\) với \(k\in Z.\)\(x=\pm\dfrac{\pi}{4}+2k\pi;x=\pm\dfrac{\pi}{6}+2k\pi\) với \(k\in Z.\)\(x=-\dfrac{\pi}{4}+k\pi;x=\dfrac{5\pi}{6}+k\pi\) với \(k\in Z.\)\(x=-\dfrac{\pi}{4}+k\pi;x=-\dfrac{\pi}{2}+2k\pi;x=\dfrac{\pi}{6}+2k\pi;x=\dfrac{5\pi}{6}+2k\pi\) với \(k\in Z.\)Hướng dẫn giải:Chuyển các tích \(\cos\frac{x}{2}\cos\frac{3x}{2}\), \(\sin\frac{x}{2}\sin\frac{3x}{2}\) thành tổng, phương trình tương đương với:
\(\cos x.\frac{1}{2}\left(\cos2x+\cos x\right)-\sin x\left[-\frac{1}{2}\left(\cos2x-\cos x\right)\right]=\frac{1}{2}\)
\(\Leftrightarrow\cos x.\cos2x+\cos^2x+\sin x\cos2x-\sin x\cos x=1\)
\(\Leftrightarrow\left(\cos x+\sin x\right).\cos2x+\cos^2x-\sin x\cos x=1\)
\(\Leftrightarrow\left(\cos x+\sin x\right).\cos2x+\left(1-\sin^2x\right)-\sin x\cos x=1\)
\(\Leftrightarrow\left(\cos x+\sin x\right).\cos2x-\sin x\left(\sin x+\cos x\right)=0\)
\(\Leftrightarrow\left(\cos x+\sin x\right).\left(\cos2x-\sin x\right)=0\)
\(\Leftrightarrow\left(\cos x+\sin x\right).\left(-2\sin^2x-\sin x+1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\sin x+\cos x=0\\-2\sin^2x-\sin x+1=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)=0\\\sin x=-1\\\sin x=\frac{1}{2}\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+\frac{\pi}{4}=k\pi\\x=-\frac{\pi}{2}+2k\pi\\x=\frac{\pi}{6}+2k\pi\\x=\pi-\frac{\pi}{6}+2k\pi\end{array}\right.\) với \(k\in Z.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{\pi}{4}+k\pi\\x=-\frac{\pi}{2}+2k\pi\\x=\frac{\pi}{6}+2k\pi\\x=\frac{5\pi}{6}+2k\pi\end{array}\right.\)với \(k\in Z.\)