Question

求2.3题解答要过程

求2.3题解答要过程解答过程画图也可以

Answer 1

(2) y=2sin^2x+2sinx-1/2，x∈[π/6,5π/6]
设t=sinx
x∈[π/6,5π/6]
t∈[sinπ/6,sinπ/2]
即：t∈[1/2,1]
y=2t^2+2t-1/2
=2(t^2+t+1/4)-2×1/4-1/2
=2(t+1/2)^2-1
t>-1/2时，y为增函数
ymin=y(1/2)
=2×(1/2+1/2)^2-1
=1
ymax=y(1)
=2×(1+1/2)^2-1
=2×9/4-1
=7/2
值域：[1,7/2]
(3) y=(cosx+5)/(2-cosx)
y(2-cosx)=cosx+5
2y-ycosx=cosx+5
2y-5=(1+y)cosx
(2y-5)/(1+y)=cosx
∵-1=<cosx<=1
∴-1=<(2y-5)/(1+y)<=1
等价于：
(2y-5)/(1+y)>=-1①
(2y-5)/(1+y)<=1②
①：(2y-5)/(1+y)+1>=0
(2y-5+1+y)/(1+y)>=0
(3y-4)/(y+1)>=0
y<=-1或者y>=4/3
②：(2y-5)/(1+y)-1<=0
[(2y-5)-(1+y)]/(1+y)<=0
(y-6)/(y+1)<=0
-1=<y<=6
值域：[4/3,6]