令2x/3=t,则f(t)=sint+cosπ/6*cost+sint*0.5=1.5sint+cosπ/6*cost=根号3*(sinπ/3*sint+cosπ/3*cost)=根号3*cos(t-π/3)
所以当t-π/3=0时,f(t)=根号3,当t-π/3=π时,f(t)=-根号3
t1=π/3 t2=4π/3
|t1-t2|=π=2(α-β)/3
所以α-β=3π/2
令2x/3=t,则f(t)=sint+cosπ/6*cost+sint*0.5=1.5sint+cosπ/6*cost=根号3*(sinπ/3*sint+cosπ/3*cost)=根号3*cos(t-π/3)
所以当t-π/3=0时,f(t)=根号3,当t-π/3=π时,f(t)=-根号3
t1=π/3 t2=4π/3
|t1-t2|=π=2(α-β)/3
所以α-β=3π/2