Tìm Min, Max của các hàm lượng giác sau:
- $\ y = \frac{{3\sin 2x + 2}}{{2{{\sin }^2}x + 2}}.$
- $\ y = sin\,x + sin\,\left( {x - \frac{\pi }{3}} \right).$
- $\ y = 2 + si{n^4}x + co{s^4}x.$
- $\ y = 1 - 3\left| {sin\,4x} \right|.$
- $\ y = \frac{{3\sin 2x + 2}}{{2{{\sin }^2}x + 2}} \Leftrightarrow 2y{\sin ^2}x + 2y = 3\sin 2x + 2$$\ \Leftrightarrow y\left( {1 - cos2x} \right) + 2y = 3\sin 2x + 2 \Leftrightarrow 3\sin 2x + y\,cos\,2x = 3y - 2\left( * \right).$ Để tồn tại x, y thì (*) phải có nghiệm hay: $\ {3^2} + {y^2} \ge {\left( {3y - 2} \right)^2} \Leftrightarrow 8{y^2} - 12y - 5 \le 0 \Leftrightarrow \left\{ \begin{array}{l}Min\,y = \frac{{3 - \sqrt {19} }}{4}\\Max\,y = \frac{{3 + \sqrt {19} }}{4}\end{array} \right.$
- $\ y = sin\,x + \,\sin \,\left( {x - \frac{\pi }{3}} \right) = 2\sin \left( {x - \frac{\pi }{6}} \right)cos\left( {\frac{\pi }{3}} \right) = \sin \left( {x - \frac{\pi }{6}} \right).$ Mà ta có: $\ - 1 \le \sin \left( {x - \frac{\pi }{6}} \right) \le 1 \Leftrightarrow \left\{ \begin{array}{l}Min\,y = - 1\\Max\,y = 1\end{array} \right..$
- $\ y = 2 + {\sin ^4}x + co{s^4}x = 2 + {\left( {{{\sin }^2}x + co{s^2}x} \right)^2} - 2{\sin ^2}xco{s^2}x = 3 - \frac{1}{2}{\sin ^2}2x.$ Mà ta có: $\ 0 \le {\sin ^2}2x \le 1 \Leftrightarrow - \frac{1}{2} \le - \frac{1}{2}{\sin ^2}2x \le 0 \Leftrightarrow \left\{ \begin{array}{l}Min\,y = 3 - \frac{1}{2} = \frac{5}{2}\\Max\,y = 3 - 0 = 3\end{array} \right.$
- Do $\ 0 \le \left| {\sin 4x} \right| \le 1 \Leftrightarrow - 3 \le - 3\left| {\sin 4x} \right| \le 0 \Leftrightarrow - 2 \le 1 - 3\left| {\sin 4x} \right| \le 1 \Leftrightarrow \left\{ \begin{array}{l}Min\,y = - 2\\Max\,y = 1\end{array} \right.$
