Question

求解该题

Answer 1

f(x)=2cos^2x+√3sin2x+a
=cos2x+1+√3sin2x+a
=2sin(2x+π/6)+1+a
2kπ-π/2<2x+π/6<2kπ+π/2
kπ-π/3<x<kπ+π/6递增

追答

在x∈[0,π/2] 2x+π/6=π/2 , x=π/6∈[0,π/2]
4=2sin(2x+π/6)+1+a
4==2+1+a
a=1

f(x)=1=2sin(2x+π/6)+3
sin(2x+π/6)=-1
2x+π/6=3π/2 , x=2π/3
2x+π/6=7π/2, x=5π/3不合题意
在x∈[0,π] ，只有x=2π/3能使等式成立，所以x的集合｛2π/3｝