Are you familiar with numerical root finding procedures?
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