f(x)=2cos^2x+根号(3)sin2x+a,x属于[0,π/2]f(x)最大值4，求满足f(x)=1x属于[-π,π]时x

f(x)=2(1/2cos^2x + √3/2sin2x)+a
=2sin(2x+π/6)+a,
当2x+π/6=π/2时,即x=π/6时,f(x)=2+a=4,a=2
f(x)=1时,sin(2x+π/6)=-1/2,f(x)周期为π,
先考虑[0,π] 此时2x+π/6∈[π/6,13π/6]
满足sin(2x+π/6)=-1/2为2x+π/6=7π/6,10π/6
解得x=π/2,5π/6,所有的解为-5π/6,-π/2,π/2,5π/6