f(x)=根号3/22sin x cos x+1/2 (1-2sin^2 x-1)-3/2
=根号3/2 sin 2x+1/2 cos 2x-1/2-3/2
=根号3/2 sin 2x+1/2 cos 2x-2
=sin 2x cos π/6+cos 2x sin π/6-2
=sin(2x+π/6)-2
因为-π/2所以-1则-3所以所求值域（-3,-3/2）