求f (x)=
|
又f (x)=
+arccos2x,知:π 2
y=
+arccos2xπ 2
∴arccos2x=y-
,π 2
x=
cos(y-1 2
)=π 2
siny,y∈[1 2
,π 2
]3π 2
故答案为:f-1(x)=
sinx,x∈[1 2
,π 2
]3π 2
求f (x)=
|
又f (x)=
+arccos2x,知:π 2
y=
+arccos2xπ 2
∴arccos2x=y-
,π 2
x=
cos(y-1 2
)=π 2
siny,y∈[1 2
,π 2
]3π 2
故答案为:f-1(x)=
sinx,x∈[1 2
,π 2
]3π 2