4sin ²2x + 8cos ²x-5=0 ai giúp e vs ạ cảm ơn

$4sin^22x+8cos^2x-5=0$

$↔ 4(1-cos^22x)+8.\dfrac{1+cos2x}{2}-5=0$

$↔ -4cos^22x+4+4(1+cos2x)-5=0$

$↔ 4cos^22x-4cos2x-3=0$

$↔ \left[ \begin{array}{l}cos2x=\dfrac{3}{2}\\cos2x=-\dfrac{1}{2}\end{array} \right.$

(Vì $cos2x∈[-1;1]$ nên loại $cos2x=\dfrac{3}{2}$)

$cos2x=-\dfrac{1}{2}$

$↔ \left[ \begin{array}{l}2x=\dfrac{2\pi}{3}+k2\pi\\2x=-\dfrac{2\pi}{3}+k2\pi\end{array} \right.$

$↔ \left[ \begin{array}{l}x=\dfrac{\pi}{3}+k\pi\\x=-\dfrac{\pi}{3}+k\pi\end{array} \right.$ $(k∈Z)$