Đáp án:
\(\left[ \begin{array}{l}x=\dfrac{\pi}{12}+\dfrac{k\pi}{2}\\x=\dfrac{-\pi}{12}+\dfrac{k\pi}{2}\end{array} \right.\)
Giải thích các bước giải:
\(2Cos4x - 8Cos2x+5=0 \\<=>2(2cos^24x-1)-8cos4x+5=0\\<=>4cos^24x-8cos4x+3=0\\<=>cos4x=\dfrac{1}{2}\\<=>\\\left[ \begin{array}{l}4x=\dfrac{\pi}{3}+k2\pi\\4x=\dfrac{-\pi}{3}+k2\pi\end{array} \right.<=>\left[ \begin{array}{l}x=\dfrac{\pi}{12}+\dfrac{k\pi}{2}\\x=\dfrac{-\pi}{12}+\dfrac{k\pi}{2}\end{array} \right.\)
