Tìm min max cua y=sinx /sinx-2cosx+3 Tìm min max cua hàm số y=2+sin2x-2√2 cos2x

a) $y = \dfrac{\sin x}{\sin x – 2\cos x + 3}$

$\Leftrightarrow y\sin x – 2y\cos x + 3y = \sin x$

$\Leftrightarrow (y – 1)\sin x – 2y\cos x + 3y = 0$

Phương trình có nghiệm

$\Leftrightarrow (y – 1)^2 + (2y)^2 \geq (3y)^2$

$\Leftrightarrow 5y^2 – 2y + 1 \geq 9y^2$

$\Leftrightarrow 4y^2 + 2y – 1 \leq 0$

$\Leftrightarrow \dfrac{-1 – \sqrt5}{4} \leq y \leq \dfrac{-1 + \sqrt5}{4}$

Vậy $\min y = \dfrac{-1 -\sqrt5}{4};\, \max y = \dfrac{-1 +\sqrt5}{4}$

b) $y = 2 + \sin2x – 2\sqrt2\cos2x$

$\Leftrightarrow \sin2x – 2\sqrt2\cos2x = y – 2$

Phương trình có nghiệm

$\Leftrightarrow 1^2 + (2\sqrt2)^2 \geq (y – 2)^2$

$\Leftrightarrow (y-2)^2 \leq 9$

$\Leftrightarrow -3 \leq y – 2 \leq 3$

$\Leftrightarrow – 1 \leq y \leq 5$

Vậy $\min y = -1;\, \max y – 5$