$ A\sin(x + a) = B\sin( x + b)$ implies $a = b$? Given $A,B,a,b$ are constants and x are variable. $A,B$ don't equal $0$.
Does $A\sin(x + a) = B\sin(x + b)$ for all $x$ implies $a = b + 2n\pi$?
I only managed to show that $a-b = n\pi$ by putting $x = -b$.

You have:


*

*taking $x = 0: A\sin a = B\sin b$

*taking $x = \frac \pi 2: A\cos a = B\cos b$


Hence $A^2 = B^2$, or $A = \pm B$. Putting in the first equation:
$$
\sin a = \sin b \implies \exists k: (a = b +  2k\pi \text{ or } 
a = \pi - b +  2k\pi )
$$
or
$$
\sin a = -\sin b \implies \exists 
k: (a = -b +  2k\pi \text{ or } 
a = \pi + b +  2k\pi )
$$
