Giải phương trình \(\cos^2\left(2x\right)=\dfrac{1}{4}\).
\(x=\pm\dfrac{\pi}{6}+k\pi,k\in Z\).\(x=\pm\dfrac{\pi}{3}+k\pi\) , \(k\in\mathbb{Z}\).\(\dfrac{\sqrt{arccos\left(\dfrac{1}{4}\right)+2k\pi}}{2},k\in Z.\)\(x=\pm\dfrac{\pi}{6}+k\pi\) , \(x=\pm\dfrac{\pi}{3}+k\pi\) , \(k\in\mathbb{Z}\).Hướng dẫn giải:\(\cos^2\left(2x\right)=\frac{1}{4}\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\cos\left(2x\right)=\frac{1}{2}\\\cos\left(2x\right)=-\frac{1}{2}\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\cos\left(2x\right)=\cos\frac{\pi}{3}\\\cos\left(2x\right)=\cos\frac{2\pi}{3}\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x=\pm\frac{\pi}{3}+2k\pi\\2x=\pm\frac{2\pi}{3}+2k\pi\end{array}\right.\) với \(k\in\mathbb{Z}\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\pm\frac{\pi}{6}+k\pi\\x=\pm\frac{\pi}{3}+k\pi\end{array}\right.\) với \(k\in\mathbb{Z}\)