Đáp án:
$S=\left\{\dfrac{k\pi}{4}\,\bigg{|}\,k\in\mathbb Z\right\}$
Giải thích các bước giải:
$\cos2x+\cos6x+4\sin^2x=2$
$⇔\cos2x+\cos6x+2(2\sin^2x-1)=0$
$⇔\cos2x+\cos6x-2\cos2x=0$
$⇔\cos6x-\cos2x=0$
$⇔\cos6x=\cos2x$
$⇔\left[ \begin{array}{l}6x=2x+k2\pi\\6x=-2x+k2\pi\end{array} \right.\,\,(k\in\mathbb Z)⇔\left[ \begin{array}{l}4x=k2\pi\\8x=k2\pi\end{array} \right.\,\,(k\in\mathbb Z)$
$⇔\left[ \begin{array}{l}x=\dfrac{k\pi}{2}\\x=\dfrac{k\pi}{4}\end{array} \right.\,\,(k\in\mathbb Z)⇔x=\dfrac{k\pi}{4}\,\,(k\in\mathbb Z)$
Vậy $S=\left\{\dfrac{k\pi}{4}\,\bigg{|}\,k\in\mathbb Z\right\}$.
