# How to solve trigonometric equations of the form $\tan(x)=m$ where $m$ is a real number?

How to solve trigonometric equations of the form $\tan(x)=m$ where $m$ is a number?

I know how to solve $\sin(x)=m$ where $m$ is a real number and $\cos(x)=m$ where $m$ is a real number but I don't know how to solve equations of the form $\tan(x)=m$.

Could you explain all things behind that equation?

• $m=\arctan(x)$. – J.R. Apr 12 '14 at 14:02
• @YourAdHere That's only 1 solution – evil999man Apr 12 '14 at 14:02
• What kind of number m is? Rational? Integer? Real? Complex? – Joe Apr 12 '14 at 14:04
• Real number yes real – user142641 Apr 12 '14 at 14:12

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 $\tan x=\overline{\rm IK}$. The points are $\rm M$ and $\rm M'$ where $\angle{\rm IOM}=\theta$ and $\angle{\rm IOM'}=\pi+\theta$. We also know that $\tan{(\theta+2k\pi)}=\tan \theta$. So the solutions are: $$x=\theta+\underbrace{2k}_{\text{even}}\pi\quad\color{grey}{\text{or}}\quad x=\pi+\theta+2k\pi=\theta+\underbrace{(2k+1)}_{\text{odd}}\pi\tag{k\in\Bbb Z}$$ You can easily see that we can also write those solutions as: $$x=\theta+n\pi\tag{n\in\Bbb Z}$$ Now it's up to you to determine that number $\theta$. You can use to achieve this task the $\arctan$ function since $\arctan(\tan\theta)=\theta$.

Side note: $\arctan$ is sometimes denoted as $\tan^{-1}$ depending on which textbook you use.

Example: Let's solve $3\tan x-\sqrt3=0\,\,(\star)$.
We have that $(\star)$ is equivalent to: $$\tan x=\dfrac{\sqrt{3}}3.$$ We also now that: $$\arctan\left(\dfrac{\sqrt{3}}3\right)=\dfrac\pi6.$$ Therefore, we conclude that: $$\text{Set of solutions to (\star)}= \left\{\dfrac\pi6+n\pi\,\left|\right.\,n\in\mathbb Z\right\}.$$

I hope this helps.
Best wishes, $\mathcal H$akim.

Notice that $\arctan:\Bbb R\rightarrow \left(-\frac\pi2,\frac\pi2\right)$ is the inverse function of $\tan$ hence $$\tan(\arctan x)=x,\;\forall x$$ and since $\tan$ is $\pi$ periodic then we have $$\tan x=m\iff x=\arctan (m)+k\pi,\quad k\in\Bbb Z$$

If $\tan x= m$, then since $$m=\frac{m}{1}=\dfrac{\left(\dfrac{\pm m}{\sqrt{m^2+1}}\right)}{\left(\dfrac{\pm 1}{\sqrt{m^2+1}}\right)},$$ then you can solve either $$\sin x =\frac{\pm m}{\sqrt{m^2+1}}$$ or $$\cos x = \frac{\pm 1}{\sqrt{m^2+1}}$$ instead, since you know how to do those.

Plot $y=\tan x$ and $y=m$ on the same graph. You will see two solutions in its fundamental period. You can easily generalize them to $n\pi + k$ where $k=\arctan m$ and $n\in \Bbb Z$

Here's an image for $m=2$. Notice that all points where $\tan x=0$ are of form $n \pi$