The Range of a particular projectile is given by $R(\theta)=151250\sin(2\theta)$ , where $R$ is measured in feet and theta is the angle of inclination of the gun barrel. Determine theta in degrees to hit a target $5000$ feet away from the gun.
So far I have this:
$$\sin(2\theta)=\frac{5000}{151250}$$ $$\sin(2\theta)=0.033058$$ Then $$2\sin(\theta)\cos(\theta) = .033058$$ $$\cos(\theta) = .033058 \vee 2\sin(\theta) = .033058$$ So $$\sin^{-1}(.016529) = .947081^{\circ} $$ $$\cos^{-1}(.033058) = 88.1056^{\circ} $$
Not sure where to go from here, think I did something wrong by splitting it up as well.
$$\sin(2\theta)=\frac{5000}{151250}$$ $$2\theta=\sin^{-1}\left(\frac{5000}{151250}\right)$$ $$\theta=\frac{1}{2}\sin^{-1}\left(\frac{5000}{151250}\right)$$