Đáp án:

\(\begin{cases}
\min y = -\sqrt2 \Leftrightarrow x = \dfrac{5\pi}{4} + k2\pi\\
\max y = \sqrt2\ \ \Leftrightarrow x = \dfrac{\pi}{4} + k2\pi
\end{cases}\quad (k\in\Bbb Z)\)

Giải thích các bước giải:

\(\begin{array}{l}
\quad y = \cos x + \cos\left(x - \dfrac{\pi}{2}\right)\\
\Leftrightarrow y = \cos x + \sin x\\
\Leftrightarrow y = \sqrt2\cos\left(x - \dfrac{\pi}{4}\right)\\
\text{Ta có:}\\
\quad -1 \leqslant \cos\left(x - \dfrac{\pi}{4}\right) \leqslant 1\\
\Leftrightarrow -\sqrt2\leqslant \sqrt2\cos\left(x - \dfrac{\pi}{4}\right) \leqslant \sqrt2\\
\Leftrightarrow -\sqrt2 \leqslant y \leqslant \sqrt2\\
\text{Do đó:}\\
+)\quad \min y = -\sqrt2\\
\Leftrightarrow \cos\left(x - \dfrac{\pi}{4}\right) = -1\\
\Leftrightarrow x - \dfrac{\pi}{4} = \pi + k2\pi\\
\Leftrightarrow x = \dfrac{5\pi}{4} + k2\pi\quad (k\in\Bbb Z)\\
+)\quad \max y = \sqrt2 \\
\Leftrightarrow \cos\left(x - \dfrac{\pi}{4}\right) = 1\\
\Leftrightarrow x  - \dfrac{\pi}{4} = k2\pi\\
\Leftrightarrow x = \dfrac{\pi}{4} + k2\pi\quad (k\in\Bbb Z)\\
\text{Vậy}\ \min y = -\sqrt2 \Leftrightarrow x = \dfrac{5\pi}{4} + k2\pi\\
\max y = \sqrt2 \Leftrightarrow x = \dfrac{\pi}{4} + k2\pi\quad (k\in\Bbb Z)
\end{array}\) 

`~rai~`

\(y=\cos x+\cos\left(x-\dfrac{\pi}{2}\right)\\\quad=\cos x+\cos\left[-\left(\dfrac{\pi}{2}-x\right)\right]\\\quad=\cos x+\cos\left(\dfrac{\pi}{2}-x\right)\\\quad=\cos x+\sin x\\\quad=\sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right).\\\text{Ta có:}-1\le\sin\left(x+\dfrac{\pi}{4}\right)\le 1\\\Leftrightarrow -\sqrt{2}\le \sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right)\le\sqrt{2}\\\Leftrightarrow -\sqrt{2}\le y\le\sqrt{2}.\\+)Min_y=-\sqrt{2}\\\Leftrightarrow \sin\left(x+\dfrac{\pi}{4}\right)=-1\\\Leftrightarrow x+\dfrac{\pi}{4}=-\dfrac{\pi}{2}+k2\pi\\\Leftrightarrow x=-\dfrac{3\pi}{4}+k2\pi.(k\in\mathbb{Z})\\+)Max_y=\sqrt{2}\\\Leftrightarrow \sin\left(x+\dfrac{\pi}{4}\right)=1\\\Leftrightarrow x+\dfrac{\pi}{4}=\dfrac{\pi}{2}+k2\pi\\\Leftrightarrow x=\dfrac{\pi}{4}+k2\pi.(k\in\mathbb{Z})\\\text{Vậy }Min_y=-\sqrt{2}\text{ khi x=}-\dfrac{3\pi}{4}+k2\pi;\\Max_y=\sqrt{2}\text{ khi x=}\dfrac{\pi}{4}+k2\pi.(k\in\mathbb{Z})\)