If sin3θsinθ=1125 then tanθ2 can have the value equal to
sin3xsinx=1125
3sinx−4sin3xsinx=1125
3−4sin2x=1125
4sin2x=6425
sin2x=1625
sinx=±45
Hence
If
sinx=45
Now
sinx=2tanx21+tan2x2
Let
tan(x2)=t
Hence
4+4t2=10t
2t2−5t+2=0
t=2 and t=12
Similarly if sinx=−45
we get
−2t2−5t−2=0
2t2+5t+2=0
t=−12 and t=−2.