(1)若x=
(2)当x∈[
解:(1)当x=
cos〈a,c〉=
=
∵0≤〈a,c〉≤π,∴〈a,c〉=
(2)f(x)=2a·b+1=2(-cos2x+sinxcosx)+1=2sinxcosx-(2cos2x-1)
=sin2x-cos2x=
∵x∈[
∴sin(2x-
∴当2x-
即x=