$\begin{array}{l} y= \sin^2x + 3\cos^2x\\ = \dfrac{1 - \cos2x}{2} + 3.\dfrac{1 + \cos2x}{2}\\ = \cos2x + 2\\ \text{Ta có:}\\ -1\leq \cos2x \leq 1\\ \Leftrightarrow 1 \leq \cos2x + 2 \leq 3\\ Hay\,\, 1 \leq y \leq 3\\ \end{array}$
Vậy $miny = 1 \Leftrightarrow \cos2x = -1 \Leftrightarrow x = \dfrac{\pi}{2} + k\pi$
$maxy = 3 \Leftrightarrow \cos2x = 1 \Leftrightarrow x = k\pi$
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$\begin{array}{l} y = \sqrt{\sin^2x+2\cos^2x}\\ =\sqrt{\dfrac{1 - \cos2x}{2} +2.\dfrac{1 + \cos2x}{2}}\\ = \sqrt{\dfrac{cos2x+3}{2}}\\ \text{Ta có:}\\ -1\leq \cos2x \leq 1\\ \Leftrightarrow 2 \leq \cos2x + 3 \leq 4\\ \Leftrightarrow 1 \leq \dfrac{cos2x+3}{2} \leq 2\\ \Leftrightarrow 1 \leq \sqrt{\dfrac{cos2x+3}{2}} \leq \sqrt2\\ Hay\,\, 1 \leq y \leq \sqrt2\\\end{array}$
Vậy $miny = 1 \Leftrightarrow \cos2x = -1 \Leftrightarrow x = \dfrac{\pi}{2} + k\pi$
$maxy = \sqrt2 \Leftrightarrow \cos2x = 1 \Leftrightarrow x = k\pi$
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$\begin{array}{l} y = \sin x +\cos x\\ = \sqrt2.\sin\left(x + \dfrac{\pi}{4}\right)\\ \text{Ta có: }\\ -1 \leq \sin\left(x + \dfrac{\pi}{4}\right) \leq 1\\ -\sqrt2 \leq \sqrt2.\sin\left(x + \dfrac{\pi}{4}\right)\leq \sqrt2\\Hay\,\,-\sqrt2 \leq y \leq \sqrt2\end{array}$
Vậy $miny = -\sqrt2 \Leftrightarrow \sin\left(x + \dfrac{\pi}{4}\right) = -1 \Leftrightarrow x = -\dfrac{3\pi}{4} + k2\pi$
$maxy = \sqrt2 \Leftrightarrow \sin\left(x + \dfrac{\pi}{4}\right) = 1 \Leftrightarrow x =\dfrac{\pi}{4} + k2\pi$
