y = 6sin2x – 8cos2x – 2 \( = 10\left( {\frac{3}{5}\sin 2x - \frac{4}{5}\cos 2x} \right) - 2\)

Đặt \(\cos \alpha = \frac{3}{5};\sin \alpha = \frac{4}{5}\)

Khi đó

y = 10(cosα sin2x – sinα cos2x) – 2 = 10sin(2x – α) – 2

Ta có: –1 ≤ sin(2x – α) ≤ 1 \( \Leftrightarrow - 10 \le 10\sin \left( {2x - \alpha } \right) \le 10 \Leftrightarrow - 12 \le y \le 8\left( {\forall x \in \mathbb{R}} \right)\)

\[Ma{x_y} = 8\] khi sin(2x – α) = 1

\( \Leftrightarrow 2x - \alpha = \frac{\pi }{2} + k2\pi \Leftrightarrow x = \frac{\pi }{4} + \frac{\alpha }{2} + k\pi \left( {k \in \mathbb{Z}} \right)\).