$\quad y = \sin x + \cos x + \sin x\cos x$
Đặt $t = \sin x + \cos x \quad \left(t\in \left[-\sqrt2;\sqrt2\right]\right)$
$\Rightarrow t^2 = 1 + 2\sin x\cos x$
$\Rightarrow \dfrac{t^2 - 1}{2} = \sin x\cos x$
Hàm số trở thành: $y = f(t) = t + \dfrac{t^2 - 1}{2}\quad \left(t\in \left[-\sqrt2;\sqrt2\right]\right)$
$\Rightarrow y' = f'(t) = 1 + t$
$y' = 0 \Leftrightarrow t= -1\in \left[-\sqrt2;\sqrt2\right]$
Bảng biến thiên:
\(\begin{array}{|c|cr|}
\hline
x & -\sqrt2 & & & -1 & & & \sqrt2\\
\hline
y' & & - & & 0 & & + &\\
\hline
&\dfrac{1- 2\sqrt2}{2}&&&&&&\dfrac{1 + 2\sqrt2}{2}\\
y & &\searrow& && & \nearrow\\
&&&&-1\\
\hline
\end{array}\)
Dựa vào bảng biến thiên, ta được:
$+)\quad \mathop{\min}y = f(-1) =-1$
$\Leftrightarrow t = -1$
$\Leftrightarrow \sin x + \cos x + -1$
$\Leftrightarrow \left[\begin{array}x =\pi + k\pi\\x = -\dfrac{\pi}{2} + k2\pi\end{array}\right.\quad (k\in \Bbb Z)$
$+)\quad \mathop{\max}y = f\left(\sqrt2\right) = \dfrac{1 + 2\sqrt2}{2}$
$\Leftrightarrow t = \sqrt2$
$\Leftrightarrow \sin x + \cos x = \sqrt2$
$\Leftrightarrow x = - \dfrac{\pi}{4} + k2\pi\quad (k\in\Bbb Z)$
