a) $y = 2 - 4cos2x$
Ta có: $-1 \leq cos2x \leq 1$
$\Leftrightarrow 4 \geq -4cos2x \geq -4$
$\Leftrightarrow 6 \geq 2 - 4cos2x \geq -2$
Vậy $y_{min} = -2; \, y_{max} = 6$
$\\$
b) $y = 3sin^22x -4$
$y = 3.\left(\dfrac{1 - cos4x}{2}\right) - 4$
$= -\dfrac{5}{2} - \dfrac{3}{2}cos4x$
Ta có: $-1 \leq cos4x \leq 1$
$\Leftrightarrow \dfrac{3}{2} \geq -\dfrac{3}{2}cos4x \geq -\dfrac{3}{2}$
$\Leftrightarrow -1 \geq -\dfrac{5}{2} - \dfrac{3}{2}cos4x \geq -4$
Vậy $y_{min} = -4; \, y_{max} = -1$
$\\$
c) $y = \sqrt{4cos^22x + 1}$
$y = \sqrt{4.\left(\dfrac{1+cos4x}{2}\right) + 1}$
$= \sqrt{2cos4x + 3}$
Ta có: $-1 \leq cos4x \leq 1$
$\Leftrightarrow -2 \leq 2cos4x \leq 2$
$\Leftrightarrow 1 \leq 2cos4x + 3 \leq 5$
$\Leftrightarrow 1 \leq \sqrt{2cos4x + 3} \leq \sqrt{5}$
Vậy $y_{min} = 1; \, y_{max} = \sqrt{5}$
$\\$
d) $y = 2|cos3x| - 4$
Ta có: $0 \leq |cos3x| \leq 1$
$\Leftrightarrow 0 \leq 2|cos3x| \leq 2$
$\Leftrightarrow -4 \leq 2|cos3x| - 4 \leq -2$
Vậy $y_{min} = -4; \, y_{max} = -2$
