Question

设函数f(x)=2根号3sinxcosx+2cos平方x-1（x属于R） 求： 若f(x0)=6/5，x0属于[π/4，π/2],求cos2x0的值

Answer 1

f(x)=2√3sinxcosx+2cosx^2-1=√3sin(2x)+cos(2x)=2sin(2x+π/6）
f(x0)=2sin(2x0+π/6)=6/5
sin(2x0+π/6)=3/5
sin2x0cosπ/6+cos2x0sinπ/6=3/5
(√3/2)sin2x0+(1/2)cos2x0=3/5
sin2x0=√[1-(cos2x0)^2]
√[1-(cos2x0)^2](√3/2)+(1/2)cos2x0=3/5
设cos2x0=t
(√3)√(1-t^2)+t=6/5=1.2
t约等于0.9929

Answer 2

f(x0)=2√3sinx0cosx0+2(cosx0)^2-1=6/5
√3sin2x0+cos2x0=6/5
√3√[1-(cos2x0)^2+cos2x0=6/5
√3√[1-(cos2x0)^2]=6/5-cos2x0
3[1-(cos2x0)^2]=[6/5-cos2x0]^2
(2cos2x0-3/5)^2=48/25
cos2x0=3/10±2√3/5
x0属于[π/4，π/2],故cos2x0=3/10-2√3/5