giải các pt 1) cos^2 2x = 1/4 2) (2 + cosx)(3cos2x+1) = 0 02/08/2021 Bởi Nevaeh giải các pt 1) cos^2 2x = 1/4 2) (2 + cosx)(3cos2x+1) = 0
1) $cos^22x=\dfrac{1}{4}$ $↔ cos2x=±\dfrac{1}{2}$ Với $cos2x=\dfrac{1}{2}$, ta có: $2x=±\dfrac{\pi}{3}+k2\pi$ $↔ x=±\dfrac{\pi}{6}+k\pi$ Với $cos2x=-\dfrac{1}{2}$, ta có: $2x=±\dfrac{2\pi}{3}+k2\pi$ $↔ x=±\dfrac{\pi}{3}+k\pi$ (Với $k∈Z$) 2) $(2+cosx)(3cos2x+1)=0$ $↔ \left[ \begin{array}{l}2+cosx=0\\3cos2x+1=0\end{array} \right.$ $→ cos2x=-\dfrac{1}{3}$ (Vì $cosx∈[-1;1]$ nên $(2+cosx)\neq0$) $↔ 2x=±arccos\Bigg(\dfrac{1}{3}\Bigg)+k2\pi$ $↔ x=±\dfrac{1}{2}arccos\Bigg(\dfrac{1}{3}\Bigg)+k\pi$ $(k∈Z)$ Bình luận
1) $cos^22x=\dfrac{1}{4}$
$↔ cos2x=±\dfrac{1}{2}$
Với $cos2x=\dfrac{1}{2}$, ta có:
$2x=±\dfrac{\pi}{3}+k2\pi$
$↔ x=±\dfrac{\pi}{6}+k\pi$
Với $cos2x=-\dfrac{1}{2}$, ta có:
$2x=±\dfrac{2\pi}{3}+k2\pi$
$↔ x=±\dfrac{\pi}{3}+k\pi$
(Với $k∈Z$)
2) $(2+cosx)(3cos2x+1)=0$
$↔ \left[ \begin{array}{l}2+cosx=0\\3cos2x+1=0\end{array} \right.$
$→ cos2x=-\dfrac{1}{3}$ (Vì $cosx∈[-1;1]$ nên $(2+cosx)\neq0$)
$↔ 2x=±arccos\Bigg(\dfrac{1}{3}\Bigg)+k2\pi$
$↔ x=±\dfrac{1}{2}arccos\Bigg(\dfrac{1}{3}\Bigg)+k\pi$ $(k∈Z)$