=2cos？ x- 1+√3sin2x+a+ 1

=cos2x+√3sin2x+a+ 1

= 2( 1/2 * cos2x+√3/2 * sin2x)+a+ 1

= 2(√3/2 * sin2x+ 1/2 * cos2x)+a+ 1

= 2(sin 2 xcos/6+cos 2 xsinπ/6)+a+ 1

=2sin(2x+π/6)+a+ 1

x∈[0，π/2]

When 2x+π/6=π/2, the value of f(x) is the largest.

That is, when sin(2x+π/6)= 1, the value of f(x) is the largest.

2+a+ 1=4

a= 1

f(x)=2sin(2x+π/6)+2

f(x)= 1

2sin(2x+π/6)+2= 1

sin(2x+π/6)=- 1/2

2x+π/6=7π/6 or 2x+π/6=-π/6.

X=π/2 or x=-π/6.