求高一三角函数的一道题解法f（x）=1/2cos平方x+根号下三/2sinxcosx+1 x属于R （ 1）求fx的最小

f（x）=1/2cos平方x+根号下三/2sinxcosx+1,x属于R,
(1)求fx的最小正周期
(2)函数在{派/12,派/4}上最大值最小值,并求出取得最大值最小值时自变量x值
(3)解不等式f（x）三次方>1/2
(1)解析：∵f(x)=1/2(cosx)^2+√3/2sinxcosx+1=1/4(cos2x+1)+√3/4sin2x+1
=1/2cos(2x-π/3)+5/4
∴fx的最小正周期为π
(2)解析：最大值点：2x-π/3=2kπ==>x=kπ+π/6
最小值点：2x-π/3=2kπ+π==>x=kπ+2π/3
∵区间[派/12,派/4]
F(π/12)=1/2cos(π/6-π/3)+5/4=√3/4+5/4
F(π/4)=1/2cos(π/2-π/3)+5/4=√3/4+5/4
∴当x=π/6时,取最大值7/4；当x=π/12或x=π/4时,取最小值√3/4+5/4；
(3)解析：设g(x)=(f(x))^3
令g’(x)=3(f(x))^2*f’(x)=0
∵f(x)>0
∴f’(x)=-sin(2x-π/3)=0
2x-π/3=2kπ==>x1=kπ+π/6
2x-π/3=2kπ+π==>x2=kπ+2π/3
f’’(x)=-2cos(2x-π/3)==> f’’(x1)0
∴g(x)在x2处取极小值g(2π/3)=(f(2π/3))^3=(3/4)^3=27/64
∵(f(x))^3>1/2==>f(x)>(1/2)^(1/3)≈0.793701
令f(x)=0.793701
cos(2x-π/3)=2*0.793701-5/2=-0.912598
2x-π/3=2kπ+0.865927π==>2x=2kπ+π/3+π-0.13407288π
==>x=kπ+2π/3-0.13407288π/2
2x-π/3=2kπ+2π-0.865927π==>2x=2kπ+2π+π/3-π+0.13407288π
==>x=kπ+2π/3+0.13407288π/2
∴不等式的解为x∈(kπ-π/3+0.06703644π,kπ+2π/3-0.06703644π)

答：
f(x)=(1/2)cos²x+(√3/2)sinxcosx+1
=(1/2)(1/2)(cos2x+1)+(√3/2)(1/2)sin2x+1
=(1/2)sin(2x+π/3)+1

（1）f(x)的最小正周期T=2π/2=π

（2）π/12<=x<=π/4，π/2<=2x+π/3<=5π/6
所以：1/2<=sin(2x+π/3)<=1
所以：1/4+1<=f(x)<=1/2+1
所以：5/4<=f(x)<=3/2
所以：
f(x)最大值为3/2，此时x=π/12
f(x)最小值为5/4，此时x=π/4

（3）f³(x)>1/2.....这个，实在不好接，运算麻烦，怎么会搞个三次方？