1) $y = 1 + \dfrac{1}{2}sinxcosx$
$= 1 + \dfrac{1}{4}(2sinxcosx)$
$= 1 + \dfrac{1}{4}sin2x$
Ta có: $-1 \leq sin2x \leq 1$
$\Leftrightarrow -\dfrac{1}{4} \leq \dfrac{1}{4}sin2x \leq \dfrac{1}{4}$
$\Leftrightarrow \dfrac{3}{4} \leq 1 + \dfrac{1}{4}sin2x \leq \dfrac{5}{4}$
Hay $\dfrac{3}{4} \leq y \leq \dfrac{5}{4}$
Vậy $miny = \dfrac{3}{4} \Leftrightarrow sin2x = -1 \Leftrightarrow -\dfrac{\pi}{4} + k\pi$
$maxy = \dfrac{5}{4} \Leftrightarrow sin2x = 1 \Leftrightarrow x = \dfrac{\pi}{4} + k\pi$ $(k \in\Bbb Z)$
2) $y = 2.cos^2x - 2$
$= 2.\dfrac{1 + cos2x}{2} - 2$
$= 1 + cos2x -2$
$= cos2x -1$
Ta có: $-1 \leq cos2x \leq 1$
$\Leftrightarrow -2 \leq cos2x -1 \leq 0$
Hay $-2 \leq y \leq 0$
Vậy $miny = -2 \Leftrightarrow cos2x = -1 \Leftrightarrow x = \dfrac{\pi}{2} + k\pi$
$maxy = 0 \Leftrightarrow cos2x = 1 \Leftrightarrow x = k\pi$ $(k \in \Bbb Z)$
3) $y = \sqrt{2 + cos^2x}$
$= \sqrt{2 + \dfrac{1 + cos2x}{2}}$
$=\sqrt{\dfrac{5 + cos2x}{2}}$
Ta có: $-1 \leq cos2x \leq 1$
$\Leftrightarrow 4 \leq 5 + cos2x \leq 6$
$\Leftrightarrow 2 \leq \dfrac{5 + cos2x}{2} \leq 3$
$\Leftrightarrow \sqrt2 \leq \sqrt{\dfrac{5 + cos2x}{2}}\leq \sqrt3$
Hay $\sqrt2 \leq y\leq \sqrt3$
Vậy $miny = \sqrt2 \Leftrightarrow cos2x = -1 \Leftrightarrow x = \dfrac{\pi}{2} + k\pi$
$maxy = \sqrt3 \Leftrightarrow cos2x = 1 \Leftrightarrow x = k\pi$ $(k \in \Bbb Z)$
4) $y = sin^4x + cos^4x$
$= (sin^2x + cos^2x)^2 - 2sin^2xcos^2x$
$= 1 - \dfrac{1}{2}(2sinxcosx)^2$
$= 1 - \dfrac{1}{2}sin^22x$
$= 1 - \dfrac{1}{4}(1 - cos4x)$
$= \dfrac{3 + cos4x}{4}$
Ta có: $-1 \leq cos4x \leq 1$
$\Leftrightarrow 2 \leq 3 + cos4x \leq 4$
$\Leftrightarrow \dfrac{1}{2} \leq \dfrac{3 + cos4x}{4} \leq 1$
Hay $\dfrac{1}{2} \leq y \leq 1$
Vậy $miny = \dfrac{1}{2} \Leftrightarrow cos4x = -1 \Leftrightarrow x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}$
$maxy = 1 \Leftrightarrow cos4x = 1 \Leftrightarrow x = k\dfrac{\pi}{2}$ $(k \in \Bbb Z)$
