Trig- question

$\tan\left(\frac{\pi}{4} \pm x\right) =\frac{\tan\left(\frac{\pi}{4}\right) \pm \tan{x}}{1 \mp \tan\left(\frac{\pi}{4}\right)\tan{x}} =\frac{1 \pm \tan{x}}{1 \mp \tan{x}}$

so ...

$\tan\left(\frac{\pi}{4} + x\right) = 3\tan\left(\frac{\pi}{4} - x\right)$

$\frac{1 + \tan{x}}{1 -\tan{x}} = \frac{3(1 - \tan{x})}{1+\tan{x}}$

$(1+\tan{x})^2 = 3(1-\tan{x})^2$

$1 + 2\tan{x} + tan^2{x}= 3 - 6\tan{x} + 3\tan^2{x}$

$0 = 2\tan^2{x} - 8 \tan{x} + 2$

$0 = \tan^2{x} - 4 \tan{x} + 1$

$0 = \tan^2{x} - 4 \tan{x} + 4 - 4 + 1$

$0 = (\tan{x} - 2)^2 - 3$

$\tan{x} = 2 \pm \sqrt{3}$

assuming $0 < x < 2\pi$ ...

$x = \arctan(2 + \sqrt{3})$

$x = \pi + \arctan(2 + \sqrt{3})$

$x = \arctan(2 - \sqrt{3})$

$x = \pi + \arctan(2 - \sqrt{3})$