I have this equation and I need to find $x$ variable:
$$1+\frac{\pi}{4}-x=\arctan x$$
can I put $\tan$ on RHS and LHS in order to find $x$:
$$\tan( 1+\frac{\pi}{4}-x)=\tan(\arctan x)$$
Secondly, after I have $x$ on RHS, how am I supposed to find $x$ if the LHS has now become a complicated expression?
On
It is certainly true that if $1+\dfrac\pi4-x=\arctan x$ then $\tan\left(1+\dfrac\pi4-x\right)=\tan(\arctan x).$
But it is not true that if $\tan\left(1+\dfrac\pi4-x\right)=\tan(\arctan x)$ then $1+\dfrac\pi4-x=\arctan x,$ since $\tan$ is not a one-to-one function.
In other words, some of the solutions of the latter equation are not solutions of the former, so you'll need to check for extraneous roots.
On
One obvious solution is following: x = 1.
If you plot the graphs of both functions: $$f(x) = 1+\frac{\pi}{4}-x$$ and $$f(x) = \arctan x$$ then it will be obvious, that except x = 1 there are no other solutions. This is the only point where intersection is possible.
For any arbitrary constants there is no analytical solution. Instead, solutions are based on some iterating converging process until you reach needed precision.
Yes, it is legal to apply tan to both parts. When two values are equal, then their tan values are equal, too. But if you are going to use that to get an equation that is easier to solve, pay attention, because another equation can have more solutions than your original equation and you will have to filter them to find out the solution of your original equation.
You are correct, but don't think this is a good strategy. Instead consider the function $$f(x)=\arctan x+x-1-\frac{\pi}{4}$$ and show that $x=1$ is the unique zero. Try to prove that $f$ is strictly increasing by considering its derivative, or just noting that $f$ is the sum of $\arctan x$ and $x$ (which are strictly increasing) plus a constant. This is more a calculus problem than a trigonometry one!