2) $y = - 2\cos^22x + \sin2x$
$\to y = -2(1 - \sin^22x) + \sin2x$
$\to y = 2\sin^22x + \sin2x - 2$
$\to y = 2\left(\sin2x + \dfrac{1}{4}\right)^2 - \dfrac{17}{8}$
Ta có:
$-1 \leq \sin2x \leq 1$
$\to - \dfrac{3}{4} \leq \sin2x + \dfrac{1}{4} \leq \dfrac{5}{4}$
$\to 0 \leq \left(\sin2x + \dfrac{1}{4}\right)^2 \leq \dfrac{25}{16}$
$\to 0 \leq 2\left(\sin2x + \dfrac{1}{4}\right)^2 \leq \dfrac{25}{8}$
$\to - \dfrac{17}{8} \leq 2\left(\sin2x + \dfrac{1}{4}\right)^2 - \dfrac{17}{8} \leq 1$
$\to - \dfrac{17}{8} \leq y \leq 1$
Vậy $\min y = - \dfrac{17}{8} \Leftrightarrow \sin2x = - \dfrac{1}{4} \Leftrightarrow \left[\begin{array}{l}x = \dfrac{1}{2}\arcsin\left(-\dfrac{1}{4}\right) + k\pi\\x = \dfrac{\pi}{2} - \dfrac{1}{2}\arcsin\left(-\dfrac{1}{4}\right) + k\pi\end{array}\right.$
$\max y = 1 \Leftrightarrow \sin2x = 1 \Leftrightarrow x = \dfrac{\pi}{4} + k\pi \quad (k \in \Bbb Z)$
$\begin{array}{l}3)\,\,\cos^23x\cos2x - \cos^2x = 0\\ \Leftrightarrow (1 + \cos6x)\cos2x - (1 + \cos2x) = 0\\ \Leftrightarrow \cos6x\cos2x - 1 = 0\\ \Leftrightarrow (4\cos^32x - 3\cos2x)\cos2x - 1 = 0\\ \Leftrightarrow 4\cos^42x - 3\cos^22x - 1 = 0\\ \Leftrightarrow \left[\begin{array}{l}\cos^22x = - \dfrac{1}{4}\quad (loại)\\\cos^22x = 1 \qquad (nhận)\end{array}\right.\\ \Leftrightarrow \cos2x = \pm 1\\ \Leftrightarrow 2x = k\pi\\ \Leftrightarrow x = k\dfrac{\pi}{2}\quad (k \in \Bbb Z)\end{array}$
