How can i find the angle X

[a*cos(b-X)]^2 = c*cos(X) where a,b,c are constant

Hey Yasir.

Are you familiar with numerical root finding procedures?

Hi NO i am not familiar with it, how ever i can look for it. Is there any other way to solve this?

Thanks

You should try expanding everything taking into account that cos(a-b) = cos(a)cos(b) - sin(a)sin(b).

You may be able to get a quadratic equation in u where u = cos(x) and then solve for u and then for x.

Alternatively, assume that [a*cos(b-X)]^2 = c*cos(X) = k. That gives 2 equations [a*cos(b-X)]^2 = k and c*cos(X) = k.
The first one gives cos(b-x ) = √k/a That is x =b - arscos√k/a
and the second one gives x = arc cos √k/c
There after find the common solution

Thanks a lot for your help. I am trying to solve this its not as simple as it looks like e.g. i am not able to find quadratic equation with only cos(x) term.