Because of sin? x+cos？ x= 1，f(x)=(2sinxcosx+sin？ x+cos？ x)+3/2/(sinx+cosx)=(sinx+cosx)？ +3/2/(sinx+cosx), let t=sinx+cosx= radical 2 sin (x+45), then 1≤t≤ radical 2, f(x)=(t? +3/2)/t = the square root of t+3/2t ≥ 6.

The equal sign holds if and only if t=3/2t, that is, the square root of t = 6/2, so the minimum value is the square root of 6.

Note that the range of X is 0 and 90, then the range of X+45 is 45 and 135, and the range of SIN (X+45) is 2/2 and 1, so 1≤t≤ 2.

Use basic inequalities. The principle is that if both A and B are positive numbers, then a+b≥2 radical ab is equal if and only if A = B ... such as widely used A? +b？ ≥2ab

Therefore, t+3/2t≥2 is simplified as the square root of t+3/2t≥6.