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.