Find all $x$ such that $\sin x = \frac{4}{5}$ and $\cos x = \frac{3}{5}$.

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 $\sin x=\frac{-4}{5}$, $\cos x=\frac{-3}{5}$.

$\cos x=\dfrac35$

$\implies x=2n\pi\pm\arccos\dfrac35$

As for $x>0,\arccos x=\arcsin\sqrt{1-x^2}=\arctan\dfrac{\sqrt{1-x^2}}x$

$\implies x=2n\pi\pm\arcsin\dfrac45$ where $n$ is any integer

Now , $\sin x=\sin\left(2n\pi\pm\arcsin\dfrac45\right)=\pm\dfrac45$

But $\sin x=+\dfrac45$

$$\implies x=2n\pi+\arcsin\dfrac45=2n\pi+\arccos\dfrac35=2n\pi+\arctan\dfrac43$$

Further if $2y=\arctan\dfrac43,\dfrac43=\tan2y=\dfrac{2\tan y}{1-\tan^2y}$

Solve the quadratic equation in $\tan y$ to find the values to be $\dfrac12,-2$

But as $0<2y<\dfrac\pi2, y=\arctan\dfrac12$

which can be validated using Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$