Đáp án:
\(\eqalign{
& \min y = - 2 \Leftrightarrow x = {{5\pi } \over 6} + k2\pi \,\,\left( {k \in Z} \right) \cr
& \max y = 2 \Leftrightarrow x = - {\pi \over 6} + k2\pi \,\,\left( {k \in Z} \right) \cr} \)
Giải thích các bước giải:
\(\eqalign{
& y = \sqrt 3 \cos x - {\mathop{\rm sinx}\nolimits} \cr
& y = 2\left( {{{\sqrt 3 } \over 2}\cos x - {1 \over 2}\sin x} \right) \cr
& y = 2\left( {\cos {\pi \over 6}\cos x - \sin {\pi \over 6}\sin x} \right)\backslash \cr
& y = 2\cos \left( {x + {\pi \over 6}} \right) \cr
& Do\,\, - 1 \le \cos \left( {x + {\pi \over 6}} \right) \le 1\,\,\forall x \in R \cr
& \Rightarrow - 2 \le 2\cos \left( {x + {\pi \over 6}} \right) \le 2\,\,\forall x \in R \cr
& \Rightarrow - 2 \le y \le 2\,\,\forall x \in R \cr
& Vay\,\,\min y = - 2 \Leftrightarrow \cos \left( {x + {\pi \over 6}} \right) = - 1 \cr
& \Leftrightarrow x + {\pi \over 6} = \pi + k2\pi \Leftrightarrow x = {{5\pi } \over 6} + k2\pi \,\,\left( {k \in Z} \right) \cr
& \max y = 2 \Leftrightarrow \cos \left( {x + {\pi \over 6}} \right) = 1 \cr
& \Leftrightarrow x + {\pi \over 6} = k2\pi \Leftrightarrow x = - {\pi \over 6} + k2\pi \,\,\left( {k \in Z} \right) \cr} \)
