Solving an Unsolvable Equation: arctan(x)+arctan(\sqrt{3}x)=\frac{7\pi}{12}

[mtayab1994]

Homework Statement

Solve the following equation:

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

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.

 

[SammyS]

Homework Statement

Solve the following equation:
$$arctan(x)+arctan(\sqrt{3}x)=\frac{7\pi}{12}$$

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.

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?

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

 

[mtayab1994]

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?

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

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

 

[SammyS]

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 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) ?

 

[mtayab1994]

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) ?

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

 

[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

 

[SammyS]

Beware of possible extraneous solutions.