f(x)=5cos²x+sin²x-4√3sinxcosx
=1+4cos²x-2√3sin(2x)
=1+2+2cos(2x)-2√3sin(2x)
=3-4[√3sin(2x)/2-cos(2x)/2]
=3-4[sin(2x)cos(π/6)-cos(2x)cos(π/6)]
=3-4sin(2x-π/6)
=3-4sin[2(x-π/12)]
T=2π/2=π
f(x)的最小正周期是π
2x-π/6=2kπ+π/2,即x=kπ+π/3,在[-π/6,π/4]上无解
2x-π/6=2kπ-π/2,即x=kπ-π/6,得x=-π/6
f(π/4)=3-2√3
f(-π/6)=7
f(x)在[-π/6,π/4]上的值域是[3-2√3,7]
或者这样
f(x)=5cos²x+sin²x-4√3sinxcosx的图象即由f(x)=-sinx的图象的横坐标缩小1倍,纵坐标扩大3倍,向右平移π/12个单位,向上平移3个单位得到
f(x)=-sin(2x)在[-π/4,π/4]单调递减
f(x)=-sin[2(x-π/12)]在[-π/6,π/3]上单调递减,即在[-π/6,π/4]上单调递减
f(π/4)=3-2√3
f(-π/6)=7
f(x)在[-π/6,π/4]上的值域是[3-2√3,7]