已知函数f(x)=sin(2x+π/6)+sin2x-π/6+2cos²x求函数的最小值及最小正周期

sinA+sinB=2sin(A+B)/2 *cos(A-B)/2
cos2a=2cosa*cosa-1
f(x)=2*sin2x*cos(π/6)+1+cos2x
=sin2x+cos2x
=√2*[sin2x*cos(π/4)+cos2x*sin(π/4)]
=√2*sin(2x+π/4)
最小值-√2
最小正周期π