Why not just solve for \(\sin(\frac{\theta}{2})\) and then use the double angle identity?
Anyway, let \(x=\frac{\theta}{2}\) for simplicity.
\(\sin(x)+\cos(x)=-\frac{6}{5}\\ \cos(x)=-\sin(x)-\frac{6}{5}\)
(by the way, it might be tempting to square both sides of \(\sin(x)+\cos(x)=-\frac{6}{5}\), and then use trig identities. Don't do that; I have made this mistake before. Unless the problem specifies that the whole thing is always nonnegative, it's better to do the way that I'm about to demonstrate.)
Substitute it into the Pythagorean identity \(\sin^2(x)+\cos^2(x)=1\):
\(\sin^2(x)+(-\sin(x)-\frac{6}{5})^2=1\\ \sin^2(x)+\sin^2(x) + \frac{12}{5}\sin(x)+\frac{36}{25}=1\\ 2\sin^2(x)+ \frac{12}{5}\sin(x)+\frac{36}{25}=1\\ 50\sin^2(x)+60\sin(x)+11=0\\\)
Quadratic formula gives:
\(\sin(x)=\frac{-6\pm\sqrt{14}}{10}\)
Now we can solve the final question using the double angle identity:
\(\sin(\theta)\\ =\sin(2x)\\ =2\cos(x)\sin(x)\\ =2(-\sin(x)-\frac{6}{5})(\sin(x))\\ =\)
Now just substitute the 2 solutions of the quadratic equation for sin(x), and you're done!
