# Arctan equation

1. Oct 9, 2012

### mtayab1994

1. The problem statement, all variables and given/known data

Solve the following equation:

$$arctan(x)+arctan(\sqrt{3}x)=\frac{7\pi}{12}$$

3. The attempt at a solution

I multiplied by tan on both sides but since we can exactly calculate tan(7pi/12) i wasn't able to get an answer. Is there something else i can do? Thank you before hand.

2. Oct 9, 2012

### SammyS

Staff Emeritus
That's not multiplying by the tangent, that's taking the tangent of both sides.

After that, do you know the angle addition identity for tangent?

$\displaystyle \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1- \tan \alpha \tan \beta}$

3. Oct 9, 2012

### mtayab1994

yes that's exactly what i did and I got:

$$1+\sqrt{3}x=tan(\frac{7\pi}{12})-\sqrt{3}x^{2}tan(\frac{7\pi}{12})$$

Should i factor out with tan on the right side to get tan(7pi/12)(1-√3x^2) or what?

I know the answer will be x=1 because arctan(1)+arctan(sqrt(3))=pi/4+pi/3=7pi/12

Last edited: Oct 9, 2012
4. Oct 9, 2012

### SammyS

Staff Emeritus
I get $(1+\sqrt{3})x=tan(\frac{7\pi}{12})-\sqrt{3}x^{2}tan(\frac{7\pi}{12})\ .$

I would divide by tan(7π/12).

Write in standard form for a quadratic equation.

Have you evaluated tan(7π/12) ?

5. Oct 9, 2012

### mtayab1994

nevermind i got it because tan(7pi/12) is tan(pi/4+pi/3)

6. Oct 9, 2012

### CAF123

Using the relation provided by SammyS, hopefully you got to $$\arctan\frac{x + \sqrt{3}x}{1-\sqrt{3}x^2} = \frac{7\pi}{12}$$

Take tangent of both sides: $\tan(\frac{7\pi}{12})$ as a sum of two angles and use the double angle relation (again) to find this.

Equate the above two and solve for x

7. Oct 9, 2012

### SammyS

Staff Emeritus
Beware of possible extraneous solutions.