Solving 0 = -v.cos(2x) + sin(x).sqrt(v^2.[sin(x)]^2 + 2ah)

[villiami]

Is it possible to make x (or a function of x) the subject of this?!?


0 = -v.cos(2x) + sin(x).sqrt(v^2.[sin(x)]^2 + 2ah) - (v^2.sin(x).[cos(x)]^2)/(sqrt(v^2.[sin(x)]^2 + 2ah))


I know its disgusting, but I would really appreciate some help.

 

[z-component]

I've converted your equation to LaTeX to help others read this monster.

Is this what you mean?

$$0 = -v cos(2x) + sin(x) \sqrt{v^{2}[sin(x)]^{2} + 2ah} - \frac{v^{2} sin(x) \cdot [cos(x)]^{2}}{\sqrt{v^{2}[sin(x)]^{2} + 2ah}}$$

 

[Jonny_trigonometry]

maybe try converting to expontials and seeing if you can simplify it more

 

[villiami]

Perfact, I hope a few ideas come out of this new form. Thanks

 

[krab]

I've converted your equation to LaTeX to help others read this monster.
Is this what you mean?
$$0 = -v cos(2x) + sin(x) \sqrt{v^{2}[sin(x)]^{2} + 2ah} - \frac{v^{2} sin(x) \cdot [cos(x)]^{2}}{\sqrt{v^{2}[sin(x)]^{2} + 2ah}}$$

Or rather:
$$0 = -v \cos(2x) + \sin(x) \sqrt{v^2\sin^2(x) + 2ah} - \frac{v^2 \sin(x) \cos^2(x)}{\sqrt{v^2\sin^2(x) + 2ah}}$$

 

[z-component]

Thanks for correcting my error, krab. :)