a) Ta có:
\(\eqalign{
& - 1 \le \cos x \le 1,\forall x \in \mathbb{R} \cr
& \Leftrightarrow 0 \le 1 + \cos x \le 2 \Leftrightarrow 0 \le 2(1 + \cos x) \le 4 \cr
& \Leftrightarrow 1 \le \sqrt {2(1 + \cos x} + 1 \le 3 \cr} \)
\(\Rightarrow y_{max}= 3\)
Dấu “ = “ xảy ra \(⇔ cos x = 1 ⇔ x = k2π (k ∈ \mathbb{Z})\)
Vậy \(y_{max}= 3\) khi \(x = k2π\)
b) Ta có:
Với mọi \(x ∈ \mathbb{R}\), ta có:
\(\eqalign{
& \sin (x - {\pi \over 6}) \le 1 \cr
& \Leftrightarrow 3\sin (x - {\pi \over 6}) \le 3 \Leftrightarrow 3\sin (x - {\pi \over 6}) - 2 \le 1 \cr
& \Leftrightarrow y \le 1 \cr} \)
Vậy \(y_{max} = 1\) \( \Leftrightarrow \sin \left( {x - \frac{\pi }{6}} \right) = 1 \)
\(\Leftrightarrow x - \frac{\pi }{6} = \frac{\pi }{2} + k2\pi \Leftrightarrow x = \frac{{2\pi }}{3} + k2\pi \,\,\left( {k \in Z} \right)\)
