Đáp án: $ \sqrt5\ge y\ge\dfrac{3\sqrt2}2$
Giải thích các bước giải:
Ta có:
$y=\sqrt{5-2\cos^2x\sin^2x}$
$\to y=\sqrt{5-\dfrac12\cdot (2\cos x\sin x)^2}$
$\to y=\sqrt{5-\dfrac12\cdot \sin^22x}$
Vì $0\sin^22x\le 1$
$\to0\le\dfrac12\sin^22x\le\dfrac12$
$\to5\ge5-\dfrac12\sin^22x\ge\dfrac92$
$\to \sqrt5\ge\sqrt{5-\dfrac12\sin^22x}\ge\dfrac{3\sqrt2}2$
$\to \sqrt5\ge y\ge\dfrac{3\sqrt2}2$
$\to GTLN_y=\sqrt5$ khi đó $\sin2x=0\to 2x=k\pi, k\in Z\to x=\dfrac12k\pi, k\in Z$
$ GTNN_y=\dfrac{3\sqrt2}2\to \sin2x=\pm1\to 2x=\dfrac12\pi+k\pi, k\in Z\to x=\dfrac14\pi+\dfrac12k\pi, k\in Z$
