Question

已知函数f(x)=2sin²(π/4+x)-√3cos2x-1,x∈R.若函数h(x)=f(x+t)的图像关于点(-π/

Answer 1

f(x)=2[sin(π/4+x)]^2-√3cos2x-1

=-cos(2x+π/2)-√3cos2x

=sin2x-√3cos2x

=2sin(2x-π/3) x∈R

所以f(x)=2sin(2x-π/3) x∈R

h(x)=f(x+t)的图象关于点（-π/6,0）对称

那么h(x-t)=f(x)的图象关于点（-π/6 + t,0）对称

而f(x)=2sin(2x-π/3)关于点（π/6,0）对称

所以-π/6+t=π/6+2kπ t∈(0,π/2)

t=π/3