How exactly do I solve arctan(3x) + arctan(x) = pi/4???

2. Originally Posted by xwrathbringerx
How exactly do I solve arctan(3x) + arctan(x) = pi/4???
You might try taking the tangent of both sides:
$\displaystyle tan(arctan(3x)+ arctan(x))= tan(\pi/4)= 1$.

On the left you will need the identity: $\displaystyle tan(a+b)= \frac{tan(a)+ tan(b)}{1- tan(a)tan(b)}$.

3. ummm does tan (a) = 3x and tan(b) = x with me eventually getting the quadratic equation 3x^2 + 4x -1 = 0?

4. Originally Posted by xwrathbringerx
How exactly do I solve arctan(3x) + arctan(x) = pi/4???
hi

$\displaystyle \tan^{-1} 3x+\tan^{-1} x=\frac{\pi}{4}$

$\displaystyle \tan^{-1} 3x=\alpha\Rightarrow \tan \alpha=3x$

$\displaystyle \tan^{-1} x=\beta \Rightarrow \tan \beta=x$

$\displaystyle \alpha+\beta=\frac{\pi}{4}$

$\displaystyle \tan (\alpha+\beta)=\tan \frac{\pi}{4}$

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

$\displaystyle \frac{3x+x}{1-3x^2}=1$

solve $\displaystyle 3x^2+4x-1=0$