Tìm max min của y=1/2sin2x+cosx mọi ng giúp e với ạ

Đáp án:

$\begin{cases}miny = -\dfrac{3\sqrt3}{4} \Leftrightarrow x = -\dfrac{5\pi}{6} + k2\pi\\maxy = \dfrac{3\sqrt3}{4} \Leftrightarrow x = \dfrac{\pi}{6} + k2\pi \quad\end{cases}\quad (k \in \Bbb Z)$

Giải thích các bước giải:

$y = \dfrac{1}{2}\sin2x + \cos x$

$= \sin x \cos x + \cos x$

Ta có:

$9 - 4\sqrt3y = 9 - 4\sqrt3(\sin x \cos x + \cos x)$

$= 9(\sin^2x + \cos^2x) - 4\sqrt3\sin x\cos x - 4\sqrt3\cos x$

$= 2(\cos^2x - 2\sqrt3\sin x\cos x + 3\cos^2x) + (4\cos^2x - 4\sqrt3\cos x + 3)$

$= 2(\cos x - \sqrt3\sin x)^2 + (2\cos x - \sqrt3)^2 \geq 0$

$\Leftrightarrow 4\sqrt3y \leq 9$

$\Leftrightarrow y \leq \dfrac{3\sqrt3}{4}$

Dấu = xảy ra $\begin{cases}\cos x - \sqrt3\sin  = 0\\2\cos x - \sqrt3 = 0\end{cases} \Leftrightarrow x = \dfrac{\pi}{6} + k2\pi \quad (k \in \Bbb Z)$

Tương tự, ta được:

$9 + 4\sqrt3y = 9 + 4\sqrt3(\sin x \cos x + \cos x)$

$= 9(\sin^2x + \cos^2x) + 4\sqrt3\sin x\cos x + 4\sqrt3\cos x$

$= 2(\cos^2x + 2\sqrt3\sin x\cos x + 3\cos^2x) + (4\cos^2x + 4\sqrt3\cos x + 3)$

$= 2(\cos x + \sqrt3\sin x)^2 + (2\cos x + \sqrt3)^2 \geq 0$

$\Leftrightarrow 4\sqrt3y \geq -9$

$\Leftrightarrow y \geq -\dfrac{3\sqrt3}{4}$

Dấu = xảy ra $\begin{cases}\cos x + \sqrt3\sin  = 0\\2\cos x + \sqrt3 = 0\end{cases} \Leftrightarrow x = -\dfrac{5\pi}{6} + k2\pi \quad (k \in \Bbb Z)$