Find all $x$ such that $\sin x = \frac{4}{5}$ and $\cos x = \frac{3}{5}$. 
Let 
  $$
\left\{ 
\begin{array}{c}
\sin x = \frac{4}{5} \\ 
\cos x = \frac{3}{5}
\end{array}
\right. 
$$
Find all of the possible values for $x$.

My try: By dividing the equations we obtain $\tan x = \frac{4}{3}$ and then $$x = \arctan\frac{4}{3} + k\pi$$ 
But WolframAlpha gives 
$$x = 2k\pi + 2\arctan\frac{1}{2}$$ 
Using $\arctan(x)+\arctan(y) = \arctan\left(\frac{x+y}{1-xy}\right)$, we get $$2\arctan\frac{1}{2} = \arctan\frac{4}{3}$$ 
but the answers are different still.
Why does this happen? And what is the correct answer?

Your original (set of) equations implies $\tan x=\frac{4}{3}$ but not the other way around. When you solve $\tan x=\frac{4}{3}$ you get the solutions of your original equation and the solutions $\sin x=\frac{-4}{5}$, $\cos x=\frac{-3}{5}$.
