问题
解答题
设向量
(Ⅰ)求f(x)最大值和此时相应的x的值; (Ⅱ)求使不等式f(x)≥
|
答案
∵f(x)=
•(a
+a
)=(sinx,cosx)•(sinx+cosx,2cosx)b
=sin2x+sinxcosx+2cos2x=1+
sin2x+1 2 1+cos2x 2
=
+3 2
(sin2x+cos2x)1 2
∴f(x)=
sin(2x+2 2
)+π 4 3 2
(I)当2x+
=π 4
π+2kπ即当x=1 2
+kπ,k∈Z时,f(x)取最大值π 8 3+ 2 2
(II)由f(x)≥
可得3 2
+3 2
sin(2x+2 2
)≥π 4 3 2
∴sin(2x+
)≥0π 4
∴2kπ≤2x+
≤2kπ+ππ 4
∴kπ-
≤ x≤kπ+π 8
,k∈Z3π 8
∴不等式的解集是{x|kπ-
≤ x≤kπ+π 8
,k∈Z}3π 8