已知f(x)=2sinxcosx+23cos2x−1−3. (1)求f(x)的最大值及此时x的值; (2)求f(x)的单调递增区间.
问题描述:
已知f(x)=2sinxcosx+2
cos2x−1−
3
.
3
(1)求f(x)的最大值及此时x的值;
(2)求f(x)的单调递增区间.
答
(1)∵2sinxcosx=sin2x,2cos2x=1+cos2x
∴f(x)=2sinxcosx+2
cos2x−1−
3
=sin2x+
3
(1+cos2x)-1-
3
=sin2x+
3
cos2x-1
3
化简,得f(x)=2sin(2x+
)-1π 3
∴当2x+
=π 3
+2kπ时,即x=π 2
+kπ(k∈Z)时,函数有最大值1π 12
(2)令−
+2kπ≤2x+π 2
≤π 3
+2kπ,得-π 2
+kπ≤x≤5π 12
+kπ,(k∈Z)π 12
∴f(x)的单调递增区间是[-
+kπ,5π 12
+kπ](k∈Z)π 12