Đáp án:
\({1 \over 2}\left[ { - \sqrt {4{a^2} + {{\left( {b + c} \right)}^2}} + b + c} \right] \le y \le {1 \over 2}\left[ {\sqrt {4{a^2} + {{\left( {b + c} \right)}^2}} + b + c} \right]\)
Giải thích các bước giải:
$$\eqalign{
& y = a\sin 2x + b{\sin ^2}x + c{\sin ^2}x \cr
& y = a\sin 2x + \left( {b + c} \right){\sin ^2}x \cr
& y = a\sin 2x + \left( {b + c} \right){{1 - \cos 2x} \over 2} \cr
& y = {1 \over 2}\left( {2a\sin 2x - \left( {b + c} \right)\cos 2x + b + c} \right) \cr
& Ta\,\,co: \cr
& - \sqrt {4{a^2} + {{\left( {b + c} \right)}^2}} \le 2a\sin 2x - \left( {b + c} \right)\cos 2x \le \sqrt {4{a^2} + {{\left( {b + c} \right)}^2}} \cr
& \Rightarrow - \sqrt {4{a^2} + {{\left( {b + c} \right)}^2}} + b + c \le 2a\sin 2x - \left( {b + c} \right)\cos 2x + b + c \le \sqrt {4{a^2} + {{\left( {b + c} \right)}^2}} + b + c \cr
& \Leftrightarrow {1 \over 2}\left[ { - \sqrt {4{a^2} + {{\left( {b + c} \right)}^2}} + b + c} \right] \le y \le {1 \over 2}\left[ {\sqrt {4{a^2} + {{\left( {b + c} \right)}^2}} + b + c} \right] \cr} $$
