Đáp án:
$2569$
Giải thích các bước giải:
$\begin{array}{l}2\cos^2x +2\cos^22x + 2\cos^23x - 3 = \cos4x(2\sin2x + 1)\\ \Leftrightarrow (1+ \cos2x) + (1 + \cos4x) + (1 + \cos6x) - 3 = 2\cos4x\sin2x + \cos4x\\ \Leftrightarrow \cos2x + \cos6x=\sin6x - \sin2x\\ \Leftrightarrow \sin2x + \cos2x = \sin6x - \cos6x\\ \Leftrightarrow \sin\left(2x + \dfrac{\pi}{4}\right) = \sin\left(6x -\dfrac{\pi}{4}\right)\\ \Leftrightarrow \left[\begin{array}{l}2x + \dfrac{\pi}{4} = 6x - \dfrac{\pi}{4} + k2\pi\\2x + \dfrac{\pi}{4} = \dfrac{5\pi}{4} - 6x + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{8} + k\dfrac{\pi}{2}\\x = \dfrac{\pi}{8} + k\dfrac{\pi}{4}\end{array}\right.\\ \Leftrightarrow x = \dfrac{\pi}{8} + k\dfrac{\pi}{4}\quad (k \in \Bbb Z)\\ \text{Ta có:}\\ \qquad 0 < x < 2018\\ \Leftrightarrow 0 < \dfrac{\pi}{8} + k\dfrac{\pi}{4} < 2018\\ \Leftrightarrow - \dfrac{\pi}{8} < k\dfrac{\pi}{4} < 2018 - \dfrac{\pi}{8}\\ \Leftrightarrow -\dfrac12 < k < \dfrac{8072}{\pi} - \dfrac12\\ \Rightarrow k = \underbrace{\left\{0;1;2;\dots;2568\right\}}_{\text{2569 giá trị k}}\\ \Rightarrow \text{2569 nghiệm thỏa mãn} \end{array}$
