Đáp án: $\dfrac{-6\sqrt{5}-9}{11}\le \:y\le \dfrac{6\sqrt{5}-9}{11}$
Giải thích các bước giải:
Ta có:
$y=\dfrac{3\sin2x-2\cos^2x}{\sin2x-2\cos2x+4}$
$\to y=\dfrac{3\sin2x-(\cos2x+1)}{\sin2x-2\cos2x+4}$
$\to y=\dfrac{3\sin2x-\cos2x-1}{\sin2x-2\cos2x+4}$
$\to 3\sin2x-\cos2x-1=y(\sin2x-2\cos2x+4)$
$\to 3\sin2x-\cos2x-1=y\sin2x-2y\cos2x+4y$
$\to 4y+1=(3-y)\sin2x+(2y-1)\cos2x$
$\to (4y+1)^2=((3-y)\sin2x+(2y-1)\cos2x)^2$
$\to (4y+1)^2\le ((3-y)^2+(2y-1)^2)(\sin^22x+\cos^22x)$
$\to (4y+1)^2\le ((3-y)^2+(2y-1)^2)\cdot 1$
$\to (4y+1)^2\le (3-y)^2+(2y-1)^2$
$\to 16y^2+8y+1\le 5y^2-10y+10$
$\to 11y^2+18y-9\le \:0$
$\to 11\left(y+\dfrac{9}{11}\right)^2-\dfrac{180}{11}\le \:0$
$\to 11\left(y+\dfrac{9}{11}\right)^2\le \dfrac{180}{11}$
$\to \dfrac{-6\sqrt{5}-9}{11}\le \:y\le \dfrac{6\sqrt{5}-9}{11}$
