Lời giải:
\(\cos ^22x+\cos 2x-\frac{3}{4}=0\)
\(\Leftrightarrow 4\cos ^22x+4\cos 2x-3=0\)
\(\Leftrightarrow (2\cos 2x+1)^2-4=0\)
\(\Leftrightarrow (2\cos 2x-1)(2\cos 2x+3)=0\)
\(\Rightarrow \left[\begin{matrix} \cos 2x=\frac{1}{2}\\ \cos 2x=\frac{-3}{2}\end{matrix}\right.\)
Nếu \(\cos 2x=\frac{1}{2}=\cos (\frac{\pi}{3})\)
\(\Rightarrow 2x=\pm \frac{\pi}{3}+2k\pi \Rightarrow x=\pm \frac{\pi}{6}+k\pi \) với $k$ nguyên
Nếu \(\cos 2x=\frac{-3}{2}\leq -1\) (vô lý- loại)
Vậy........
cos2 2x +cos2x - \(\frac{3}{4}\) = 0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}cos2x=\frac{1}{2}\left(N\right)\\cos2x=\frac{-3}{2}\left(L\right)\end{matrix}\right.\)
* cos2x=\(\frac{1}{2}\)
cos 2x=cos \(\frac{\pi}{3}\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x=\frac{\pi}{3}+k2\pi\\2x=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=\frac{\pi}{6}+k\pi\\x=-\frac{\pi}{6}+k\pi\end{matrix}\right.\)(k\(\in\)Z)
