Can you solve this (hard equation) for G?

Question

x = 2*arccos((G-3)/3) - sin(2*arccos((G-3)/3)

if it helps, G must be between 0 and 6. Any help would be greatly appreciated.

so with help from sudhakarbabu below, I got:

x = pi - 2arcsin((G-3)/3) - 2((G-3)/3)*sqrt(1-((G-3)/3)^2).

Can we simplify this so we have G on one side of the equation and a function of x on the other?

thanks a lot for the help

Answer 1

Hi,

There are two unknowns x and G,with only one equation.
However the equation is simplified as follows;

x = 2*arccos((G-3)/3) - sin(2*arccos((G-3)/3)
Let (G-3)/3 = D, for ease of writing the eqns.
x = 2*arccosD - sin(2*arccosD)
x = 2*arccosD - sin[2*{(pi/2) - arc sinD}]
x = 2*arccosD - sin[2*{(pi/2) - arc sinD}]
x = 2*arccosD - sin[pi - 2*arcsinD}]
x = 2*arccosD - sin[2*arcsinD]
x = 2[(pi/2) - arcsinD] - 2[sin(arcsinD)][cos(arcsinD)]
x = pi - 2arcsinD - 2[sin(arcsinD)][cos(arcsinD)]
x = pi - 2arcsinD - 2D[cos(arcsinD)]
x = pi - 2arcsinD - 2D[1 - D^2]^0.5

You can further take off from here after correcting the initial equation if necessary.

Good luck

Answer 2

You won't be able to find an analytic expression for this because the equation is transcendental in G. If you substitute y = acos((G-3)/3) then your equation becomes

x = 2y - sin(2y) with y between 0 and pi

You could find a local inverse of this function by using a taylor series expansion for sin(2y) and then solving for y(x) but this will only be a true near the point that you linearize about.

Answer 3

Well first and foremost you can not solve for G and get a numerical answer because you have two unknowns in your equation - G and X.
If you solve for G you will get
G= (3X+6arccos-6arccos(sin))/(2arccos-2arccos(sin))

Answer 4

What is x? You can't have 0 < G < 6 is x isn't specified.