<s>[INST] Solving $d = \frac{v^2}{2g} \left( 1+ \sqrt{ 1+ \frac{2g y_0}{v^2 \sin^2{\theta}} } \right)\sin{2\theta}$ for $\theta$ 
$$d = \frac{v^2}{2g} \left( 1+ \sqrt{  1+ \frac{2g y_0}{v^2 \sin^2{\theta}}  } \right)\sin{2\theta}$$

Need to find an equation for the angle.
This is the equation for the range of a projectile thrown from a height at an angle. I have tried to solve it myself and even used sites like Wolfram Alpha and Symbolab but to no avail. Could anybody please help?
The equation and derivation for the range of projectile launched at an angle from a height
 [/INST] After a little messing around, I figured it out. Here it goes:
Given equation:
$$d = \frac{v^2}{2g} \left( 1+ \sqrt{  1+ \frac{2g y_0}{v^2 \sin^2{\theta}}  } \right)\sin{2\theta}$$
$$\implies \dfrac{2gd}{v^2}=(\sin 2\theta) +\left[\dfrac{(2\sin\theta\cos\theta)\sqrt{v^2\sin^2\theta+2gy_0}}{v\sin\theta}\right]$$
$$\implies \dfrac{2gd}{v^2}=2\cos\theta\left( \sin\theta+\sqrt{\sin^2\theta+\frac{2gy_0}{v^2}} \right)$$
$$\implies \dfrac{gd}{v^2}\sec\theta-\sin\theta=\sqrt{\sin^2\theta+\frac{2gy_0}{v^2}}$$
Let $\dfrac{gd}{v^2}=a$ and $\dfrac{2gy_0}{v^2}=b.$
$$\implies a\sec\theta-\sin\theta=\sqrt{\sin^2\theta+b}$$
Squaring on both sides,
$$\implies a^2\sec^2\theta+\sin^2\theta-2a\sec\theta\sin\theta=\sin^2
\theta+b$$
$$\implies a^2(1+\tan^2\theta)-2a\tan\theta-b=0$$
$$\implies a^2\tan^2\theta-2a\tan\theta+(a^2-b)=0$$
Can you take it from here?
</s>