Đáp án: $\left[\begin{array}{l} x = \pm\dfrac{\pi}{3} + k2\pi\\x = \pm\dfrac{\pi}{6}+ k2\pi\end{array}\right.\quad (k \in \Bbb Z)$ Giải thích các bước giải: $2\cos2x – 2(\sqrt3 + 1)\cos x +2 + \sqrt3 =0$ $\Leftrightarrow 2(2\cos^2x – 1) -2(\sqrt3 + 1)\cos x +2 + \sqrt3 =0$ $\Leftrightarrow 4\cos^2x -2(\sqrt3 + 1)\cos x + \sqrt3 =0$ $\Leftrightarrow \left[\begin{array}{l}\cos x = \dfrac{1}{2}\\\cos x = \dfrac{\sqrt3}{2}\end{array}\right.$ $\Leftrightarrow \left[\begin{array}{l} x = \pm\dfrac{\pi}{3} + k2\pi\\x = \pm\dfrac{\pi}{6}+ k2\pi\end{array}\right.\quad (k \in \Bbb Z)$ Reply
Đáp án:
$\left[\begin{array}{l} x = \pm\dfrac{\pi}{3} + k2\pi\\x = \pm\dfrac{\pi}{6}+ k2\pi\end{array}\right.\quad (k \in \Bbb Z)$
Giải thích các bước giải:
$2\cos2x – 2(\sqrt3 + 1)\cos x +2 + \sqrt3 =0$
$\Leftrightarrow 2(2\cos^2x – 1) -2(\sqrt3 + 1)\cos x +2 + \sqrt3 =0$
$\Leftrightarrow 4\cos^2x -2(\sqrt3 + 1)\cos x + \sqrt3 =0$
$\Leftrightarrow \left[\begin{array}{l}\cos x = \dfrac{1}{2}\\\cos x = \dfrac{\sqrt3}{2}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l} x = \pm\dfrac{\pi}{3} + k2\pi\\x = \pm\dfrac{\pi}{6}+ k2\pi\end{array}\right.\quad (k \in \Bbb Z)$