Tìm GTLN, GTNN của hàm số y = sin²x – cos²x + 2sinxcosx + 5.

y = sin²x – cos²x + 2sinxcosx + 5

= sin2x – cos2x + 5

= \(\sqrt 2 \sin \left( {2x - \frac{\pi }{4}} \right) + 5\)

Ta có: –1 ≤ \(\sin \left( {2x - \frac{\pi }{4}} \right)\) ≤ 1

⇒ \( - \sqrt 2 \le \sqrt 2 \sin \left( {2x - \frac{\pi }{4}} \right) \le \sqrt 2 \)

⇒ \(5 - \sqrt 2 \le \sqrt 2 \sin \left( {2x - \frac{\pi }{4}} \right) + 5 \le 5 + \sqrt 2 \)

Hay \(5 - \sqrt 2 \le y \le 5 + \sqrt 2 \)

Vậy max y = \(5 + \sqrt 2 \) khi \(2x - \frac{\pi }{4} = \frac{\pi }{2} + k2\pi \Leftrightarrow x = \frac{{3\pi }}{8} + k\pi \left( {k \in \mathbb{Z}} \right)\)

Min y = \(5 - \sqrt 2 \)khi \(2x - \frac{\pi }{4} = - \frac{\pi }{2} + k2\pi \Leftrightarrow x = - \frac{\pi }{8} + k\pi \left( {k \in \mathbb{Z}} \right)\).