tangent of a square root function I'm asked to solve for x:  $\tan (\sqrt{2x+3}) =3$
I put my calculator in radian mode, then took inverse tan of $3 = 1.25$ (approx).
Squaring both sides, I then get $2x+3=1.56$, so $x=-0.72$.  
Two questions, (a) does squaring both sides create possible non-answers?  (b) Since tangent is cyclical, should my answer be $x=-0.72 + 2\pi n$?
 A: First, note that the period of the tangent function is $\pi$. Therefore, if $t=\tan(\theta)$, then $t=\tan(\theta +n\pi)$ for all integer values of $n$ and $\theta = \arctan(t)-n\pi$ for all integer values of $n$.
However, if the values of $\theta$ are restricted, say to non-negative real numbers, then the permissible values of $n$ will be restricted accordingly.
For the problem of interest, we have
$$\tan(\sqrt{2x+3})=3 \tag 1$$
From $(1)$, we find that $\sqrt{2x+3}=\arctan(3)+n\pi$ for $n\ge 0$, where the arctangent function is taken on the principle branch for which $|\arctan(x)|<\pi/2$.  Note that since $\sqrt{x+3}\ge 0$, the permissible values of $n$ are indeed the non-negative integers.
Next, we solve for $x$ to find
$$\bbox[5px,border:2px solid #C0A000]{x=\frac12\left(\left(\arctan(3)+n\pi\right)^2-3\right)}$$
for all $n\ge 0$. There are no spurious solutions introduced upon solving for $x$ here.
Note that the values of $x$ are not periodic since the term $n\pi$ appears inside the squaring operator.
