trigo function

Let t = tan(x/2), therefore, sin x = 2t/(1+t^2) and cos x = (1-t^2)/(1+ t^2).Sub. into the function, we get
y = [3 - 2t/(1+t^2)]/[2 + (1-t^2)/(1+t^2)] = [3(1+t^2) - 2t]/[2(1 +t^2) + ( 1 - t^2)]
= (3t^2 - 2t +3)/(2 + 2t^2 + 1 - t^2) = (3t^2 - 2t + 3)/(t^2 + 3).
That is
y(t^2 + 3) = 3t^2 - 2t + 3
yt^2 + 3y - 3t^2 + 2t - 3 = 0
(y-3)t^2 + 2t +(3y - 3) = 0.
For t to be real, delta > or equal to 0. That is
4 - 4(y-3)(3y -3) > = 0
1 - 3y^2 + 12y - 9 > = 0
3y^2 - 12y + 8 < = 0
So y is between [12 +/- sqrt(144 - 96)]/6, that is 2 +/-( 2/3)sqrt3.
So y max. is 2 + (2/3)sqrt 3 and
y min. = 2 - (2/3) sqrt 3.