我剛答過的題:
(I)
f(x)=3sin^2x+2√3sinxcosx+5cos^2x
=sin^2 x+2√3sinxcosx+3cos^2 x+2(sin^2 x+cos^2 x)
=sin^2 x+2√3sinxcosx+(√3)^2·cos^2 x+2
=(sinx+√3cosx)^2+2
=[√(1^2+(√3)^2)·sin(x+arctan(√3/1)]^2+2
=[2sin(x+π/3)]^2+2
=4·sin^2 (x+π/3) +2
∵sin(x+π/3)≤1
∴sin^2 (x+π/3)≤1
∴f(x)=4·sin^2 (x+π/3) +2≤6;最大值是6.
f(x)=4·sin^2 (x+π/3) +2
=2·[2sin^2 (x+π/3)-1] +2 +2
=-2·[1-2sin^2 (x+π/3)]+4
=-2cos(2x+3π/3)+4
則f（x）的最小正周期是2π/2=π.
(II)
代入原式:
f(a)=3sin^2 a+2√3sinacosa+5cos^2 a
=5;
∴3sin^2 a+2√3sinacosa=5(1-cos^2 a)=5sin^2 a;
2√3sinacosa=2sin^2 a;
√3sinacosa=sin^2 a;
兩邊同時除以sin^2 a得
√3/tana=1;
∴tana=√3.