Tìm GTLN, GTNN của hàm số y = –sin²x – cosx + 2.

y = –sin²x – cosx + 2

= 1 – sin²x – cosx + 1

= cos²x – cosx + 1

= \({\left( {\cos x - \frac{1}{2}} \right)^2} + \frac{3}{4}\)

Vì –1 ≤ cosx ≤ 1

Nên: \(\frac{{ - 3}}{2} \le \cos x - \frac{1}{2} \le \frac{1}{2}\)

Suy ra: \(\frac{9}{4} \ge {\left( {\cos x - \frac{1}{2}} \right)^2} \ge \frac{1}{4}\)

⇔ \(3 \ge {\left( {\cos x - \frac{1}{2}} \right)^2} + \frac{3}{4} \ge 1\)

Vậy max y = 3 khi cosx = 1 ⇔ x = k2π (k ∈ ℤ).

Min y = 1 khi cosx = –1 ⇔ x = π + k2π (k ∈ ℤ).