已知函数f(x)=2sinxcosx+sin(2x+
(1)若x∈R,求f(x)的最小正周期和单调递增区间; (2)设x∈[0,
|
(1)f(x)=sin2x+cos2x=
sin(2x+2
)π 4
周期T=
=π;2π 2
令2kπ-
≤2x+π 2
≤π 4
+2kπ,得kπ-π 2
≤x≤kπ+3π 8 π 8
所以,单调递增区间为[kπ-
,kπ+3π 8
],k∈Zπ 8
(2)若0≤x≤
,则π 3
≤2x+π 4
≤π 4
,sin11π 12
=sin11π 12
=sin(π 12
-π 4
)=π 6
<sin
-6 2 4
∴π 4
≤sin(2x+
-6 2 4
)≤1,π 4
≤
-13 2
sin(2x+2
)≤π 4 2
即f(x)的值域为[
,
-13 2
]2