Solve θ=2xarctan(41/8v0)

$$\theta = 2 x \arctan ( \frac { 41 } { 8 v 0 } )$$

Solve for x

$$x=\frac{\theta }{2\arctan(\frac{41}{8v_{0}})}$$

$$v_{0}\neq 0$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$2x\arctan(\frac{41}{8v_{0}})=\theta$$

The equation is in standard form.

$$2\arctan(\frac{41}{8v_{0}})x=\theta$$

Divide both sides by 2\arctan(\frac{41}{8}v_{0}^{-1}).

$$\frac{2\arctan(\frac{41}{8v_{0}})x}{2\arctan(\frac{41}{8v_{0}})}=\frac{\theta }{2\arctan(\frac{41}{8v_{0}})}$$

Dividing by 2\arctan(\frac{41}{8}v_{0}^{-1}) undoes the multiplication by 2\arctan(\frac{41}{8}v_{0}^{-1}).

$$x=\frac{\theta }{2\arctan(\frac{41}{8v_{0}})}$$

Solve for v_0

$$\left\{\begin{matrix}v_{0}=\frac{41\cot(\frac{\theta }{2x})}{8}\text{, }&\nexists n_{2}\in \mathrm{Z}\text{ : }\theta =2\pi n_{2}x+\pi x\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }\theta =2\pi n_{1}x\text{ and }\theta >-\pi x\text{ and }\theta <\pi x\text{ and }x>0\\v_{0}\neq 0\text{, }&\theta =0\text{ and }x=0\end{matrix}\right.$$