Question

已知函数∫（x）=2√3sinxcosx+2cos2x-1（x∈R）

（1）求函数∫（x）的最小正周期及在区间[0，π/2]上的最大值和最小值；
（2）若∫（x0）=6/5，x0∈[π/4，π/2]，求cos2x0的值

Answer 1

1
f（x）=2√3sinxcosx+2cos2x-1
=√3sin2x+cos2x
=2sin(2x+π/6)
最小正周期T=2π/2=π
∵x∈[0，π/2]∴2x+π/6∈[π/6,7π/6]
∴2x+π/6=π/2时，f(x)max=2
2x+π/6=7π/6时，f(x)min=-1
2
f(x0)=6/5 ,即 2sin(2x0+π/6)=6/5
∴sin(2x0+π/6)=3/5
∵ x0∈[π/4，π/2]，
∴2x0+π/6∈[2π/3,7π/6]
∴cos(2x0+π/6)=-4/5
∴cos2x0=cos[(2x0+π/6)-π/6]
=cos(2x0+π/6)cosπ/6+sin(2x0+π/6)sinπ/6
=-4/5×√3/2+3/5×1/2=(3-4√3)/10