Đáp án:
1.3b) $\begin{cases}miny = -\sqrt3 \Leftrightarrow x = \dfrac{7\pi}{6} + k2\pi\\maxy = \sqrt3 \Leftrightarrow x = \dfrac{\pi}{6} + k2\pi\end{cases}\,\,\,\,\,(k \in \Bbb Z)$
1.3d) $\begin{cases}miny = \dfrac{3\sqrt2}{2} \Leftrightarrow x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}\\
maxy = \sqrt5 \Leftrightarrow x = k\dfrac{\pi}{2}\end{cases}\,\,\,\,\,(k \in \Bbb Z)$
Giải thích các bước giải:
$\begin{array}{l}1.3b)\,\,y = \cos x + \cos\left(x - \dfrac{\pi}{3}\right)\\ = 2\cos\left(x - \dfrac{\pi}{6}\right).\cos\dfrac{\pi}{6}\\ =\sqrt3\cos\left(x - \dfrac{\pi}{6}\right)\\ \text{Ta có:}\\ -1 \leq \cos\left(x - \dfrac{\pi}{6}\right)\leq 1\\ \Leftrightarrow -\sqrt3 \leq \cos\left(x - \dfrac{\pi}{6}\right) \leq \sqrt3\\ Hay\,\,-\sqrt3 \leq y \leq \sqrt3\\ Vậy\,\,miny = -\sqrt3 \Leftrightarrow \cos\left(x - \dfrac{\pi}{6}\right) = -1\Leftrightarrow x = \dfrac{7\pi}{6} + k2\pi\\ maxy = \sqrt3 \Leftrightarrow \cos\left(x - \dfrac{\pi}{6}\right) = 1 \Leftrightarrow x = \dfrac{\pi}{6} + k2\pi&&(k \in \Bbb Z)\\ 1.3d) \,\,y=\sqrt{5 - 2\cos^2x\sin^2x}\\ = \sqrt{5 - \dfrac{1}{2}.(2\cos x.\sin x)^2}\\ = \sqrt{5 - \dfrac{1}{2}\sin^22x}\\ =\sqrt{5 - \dfrac{1}{4}(1 - \cos4x)}\\ = \dfrac{\sqrt{19+\cos4x}}{2}\\ \text{Ta có:}\\ -1 \leq \cos4x \leq 1\\ \Leftrightarrow 18 \leq 19 + \cos4x \leq 20\\ \Leftrightarrow 3\sqrt2 \leq \sqrt{19+\cos4x} \leq 2\sqrt5\\ \Leftrightarrow \dfrac{3\sqrt2}{2} \leq \dfrac{\sqrt{19+\cos4x}}{2} \leq \sqrt5\\ Hay\,\, \dfrac{3\sqrt2}{2} \leq y \leq \sqrt5\\ Vậy\,\,miny = \dfrac{3\sqrt2}{2} \Leftrightarrow \cos4x = -1 \Leftrightarrow x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}\\ maxy = \sqrt5 \Leftrightarrow \cos4x = 1 \Leftrightarrow x = k\dfrac{\pi}{2}&&(k \in \Bbb Z)\end{array}$
