For each integer n, \tan x is a bijection from (\tfrac{2n-1}{2}\pi,\tfrac{2n+1}{2}\pi) to \mathbb{R}. So for each positive integer n, there is exactly one positive solution to x = \tan x ...

In the particular case b=1, it is known that the roots of \tan(x_n)=x_n are close to : x_n=(n-\frac{1}{2})\pi-\frac{1}{(n-\frac{1}{2})\pi}-\frac{2}{3}(\frac{1}{(n-\frac{1}{2})\pi})^3-\frac{13}{15}(\frac{1}{(n-\frac{1}{2})\pi})^5-\frac{146}{105}(\frac{1}{(n-\frac{1}{2})\pi})^7-\frac{781}{315}(\frac{1}{(n-\frac{1}{2})\pi})^9-... ...

If \tan(x)=3, then \tan^2(x)=9. This means that \frac{\sin^2x}{1-\sin^2x}=9. So, \sin^2(x)=\frac9{10}; in other words (at least if we're on the first quadrant), \sin(x)=\frac3{\sqrt{10}}. ...

\displaystyle{x}={1.3528},{4.4674} Explanation: To solve use, use the inverse tangent function: \displaystyle{\tan{{\left({x}\right)}}}={4} \displaystyle\Rightarrow{x}={\arctan{{\left({4}\right)}}}={1.3258} ...

\displaystyle{78}^{\circ}{69}+{k}{180}^{\circ} Explanation: Calculator and unit circle --> tan x = 5 --> arc \displaystyle{x}={78}^{\circ}{69}+{k}{180}^{\circ}

If (\mathscr C) is a unit circle, then \overline{\rm IK}=\tan \theta (see the diagram). \phantom{X} From this picture we can see that there are two points in the unit circle that satisfy \tan x=\overline{\rm IK} ...