f(x)=2cos²x+2√3sinxcosx+a
=cos2x+1+√3sin2x+a
=2×[(1/2)×cos2x+(√3/2)×sin2x]+a+1
=2sin(2x+π/6)+(a+1)
(1)
2kπ-π/2≤2x+π/6≤2kπ+π/2时,f(x)单调递增
解得,kπ-π/3≤x≤kπ+π/6
2kπ+π/2≤2x+π/6≤2kπ+3π/2时,f(x)单调递减
解得,kπ+π/6≤x≤kπ+2π/3
所以,f(x)的单调增区间为[kπ-π/3,kπ+π/6]
单调减区间为[kπ+π/6,kπ+2π/3]
(其中,k为整数)
(2)
0≤x≤π/2时
π/6≤2x+π/6≤7π/6
当 2x+π/6=π/2,
即 x=π/6时
f(x)有最大值=2+(a+1)=3+a
因为f(x)的最大值为4
即,a+3=4
解得,a=1
所以,a的值为1