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?
Best Answer
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.