sin(2a-b)=3/5,sinb=-12/13,a∈(π/2,π),b∈(-π/2,0),求sina的值

sinb=-12/13
∵b∈(-π/2,0)
∴cosb>0
cosb=5/13
sin(2a-b)=3/5
sin2acosb-cos2asinb=3/5
5sin2a/13+12cos2a/13=3/5
5sin2a+12cos2a=39/5
10sinacosa+12cos^2a-12sin^2a=39/5
设sina=t
10t√1-t^2+12(1-2t^2)=39/5
100t^2-100t^4=1521/25+144(1-2t^2)^2-936/5(1-2t^2)
...............(过程复杂)
整理得:676t^4-1508/5t^2+441/25=0
delta=1081600/25
[t^2]1=548/6760
[t^2]2=2548/6760
∵a∈(π/2,π)
∴sina>0
t1=√548/6760,t2=-√548/6760(舍)
t3=√2548/6760,t4=-√2548/6760(舍)
t1=√548/6760=7(√1.3)/13
t3=√2548/6760=(√13.7)/13
∴sina1=7(√1.3)/13,sina2=(√13.7)/13