不等式化为：t≥(1-cos^2B)/(2+cosB)
设x=cosB
y=(1-x^2)/(2+x)
x^2+yx+2y-1=0
∆=y^2-4(2y-1)≥0
y≥4+2√3,或y≤4-2√3
又x∈[0,1]
f(0)f(1)≤0
(2y-1)((1+y+2y-1)≤0
0≤y≤1/2
所以0≤y≤1/2
t≥1/2