∵a•b=√3sinx cosx+ cos^2x=(√3/2)sin2x+[(2 cos^2x-1)+1]/2=(√3/2)sin2x+(1/2)cos2x+(1/2)= sin[2x+(π/6)]+ (1/2)=1则,sin[2x+(π/6)]= 1/2=sin(π/6)又∵x∈[-π/4,π/4]∴2x+(π/6)= π/6∴x=0（2)又(1...