Question

求解一道数学题已知函数f(x)=√3*sinwx*coswx+cos²wx-1/2(w>0),其最小正周期为π

Answer 1

解析：∵f(x)=√3sinwxcoswx+cos²wx-1/2(w>0)
=(√3/2)sin2wx+(1/2)cos2wx
=sin(2wx+π/6)
∵f(x)最小正周期T=2π/(2w)=π/2，w=2
∴f(x)=sin(4x+π/6)
将f(x)向右平移π/8个单位得到：F(x)=sin[4(x-π/8)+π/6)]=sin(4x-π/3)
再将横坐标伸长到原来的2倍得到：y=g(x)=sin(2x-π/3)
∵方程g(x)+k=0在[0,哗场糕渡蕹盗革醛宫互π/2]仅有一个实数解
sin(2x-π/3)+k=0
∵g(0)=sin(-π/3)=-√3/2，g(π/2)=sin(π-π/3)=√3/2
∴k=-sin(2x-π/3)==>-√3/2