Đáp án:
$\left[\begin{array}{l}x=\pm\dfrac{\pi}{9}+k\dfrac{2\pi}{3}\\x = \pm\dfrac{\pi}{18}+k\dfrac{2\pi}{3}\end{array}\right.\quad(k \in \Bbb Z)$
Giải thích các bước giải:
$\begin{array}{l}4\sin^23x+2(\sqrt3+1)\cos3x-\sqrt3=4\\ \Leftrightarrow 4(1 - \cos^23x) + 2(\sqrt3+1)\cos3x-\sqrt3=4\\ \Leftrightarrow -4\cos^23x + 2(\sqrt3+1)\cos3x - \sqrt3 = 0\\ \Leftrightarrow \left[\begin{array}{l}\cos3x=\dfrac{1}{2}\\\cos3x = \dfrac{\sqrt3}{2}\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}3x=\pm\dfrac{\pi}{3}+k2\pi\\3x = \pm\dfrac{\pi}{6}+k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x=\pm\dfrac{\pi}{9}+k\dfrac{2\pi}{3}\\x = \pm\dfrac{\pi}{18}+k\dfrac{2\pi}{3}\end{array}\right.&(k \in \Bbb Z)\end{array}$
