Đáp án:
$S=\left\{\dfrac{\pi}{16}+k\dfrac{\pi}{2};-\dfrac{\pi}{16}+k\dfrac{\pi}{2};\dfrac{3\pi}{16}+k\dfrac{\pi}{2};-\dfrac{3\pi}{16}+k\dfrac{\pi}{2}\,\bigg{|}\,k\in\mathbb Z\right\}$
Giải thích các bước giải:
$1-2\cos^24x=0$
$⇔2\cos^24x=1$
$⇔\cos^24x=\dfrac{1}{2}$
$⇔\cos4x=±\dfrac{\sqrt 2}{2}$
TH1: $\cos4x=\dfrac{\sqrt 2}{2}$
$⇔\left[ \begin{array}{l}4x=\dfrac{\pi}{4}+k2\pi\\4x=-\dfrac{\pi}{4}+k2\pi\end{array} \right.\,\,(k\in\mathbb Z)$
$⇔\left[ \begin{array}{l}x=\dfrac{\pi}{16}+k\dfrac{\pi}{2}\\x=-\dfrac{\pi}{16}+k\dfrac{\pi}{2}\end{array} \right.\,\,(k\in\mathbb Z)$
TH2: $\cos4x=-\dfrac{\sqrt 2}{2}$
$⇔\left[ \begin{array}{l}4x=\dfrac{3\pi}{4}+k2\pi\\4x=-\dfrac{3\pi}{4}+k2\pi\end{array} \right.\,\,(k\in\mathbb Z)$
$⇔\left[ \begin{array}{l}x=\dfrac{3\pi}{16}+k\dfrac{\pi}{2}\\x=-\dfrac{3\pi}{16}+k\dfrac{\pi}{2}\end{array} \right.\,\,(k\in\mathbb Z)$
Vậy $S=\left\{\dfrac{\pi}{16}+k\dfrac{\pi}{2};-\dfrac{\pi}{16}+k\dfrac{\pi}{2};\dfrac{3\pi}{16}+k\dfrac{\pi}{2};-\dfrac{3\pi}{16}+k\dfrac{\pi}{2}\,\bigg{|}\,k\in\mathbb Z\right\}$.
