Where I am wrong and why ?

How to solve it then ?

Ah well, you can’t write ACB and ABC as x

ACB = ABC but they are not equal to AOC

So, take COA = CAO =x

and CBA =ACB = y

But ACB (and ABC) can be considered as 2x instead of y since it is the external angle for the left triangle.

If you’re solving using trigno, it becomes 2.(5cos(x)) = 5 + 2.(5cos(2x))

[obtained by equating Radius (OA) = Radius (OB)]

If solving without trigno, use OAB = ABC, and write both in terms of x