设f(x)=6cos^2x-根3sin2x
问题描述：
设f(x)=6cos^2x-根3sin2x
1·求f(x)的最大值及最小正周期
2·若锐角a满足f(a)=3-2根3,求tan4\5a
答
f(x)=6cos^2x-√3sin2x
=3(1+cos2x)-√3sin2x
=3+2√3(cosπ/6cos2x-sinπ/6sin2x)
=3+2√3cos(2x+π/6)
fmax(x)=3+2√3,
T=2π/2=π
f(a)=fmin(x),
2a+π/6=π,
a=π/2-π/12=5π/12,
4a/5=π/3,
tan4/5a=tanπ/3=√3 .