Question

Solve the following equations.
$\sqrt{3}-tanx=tan(\frac{\pi }{4}-x).$

Answer 1

$\sqrt{3}-\tan x=\tan (\frac{\pi }{4}-x)$

$\Rightarrow \sqrt{3}-\tan x=\frac{1-\tan x}{1+\tan x}$

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

${\tan }^{2}x-(1+2\sqrt{3})\tan x+(1-\sqrt{3})=0$

$\tan x=\frac{1+2\sqrt{3}}{2}+\frac{\sqrt{{(1+2\sqrt{3})}^2-4(1-\sqrt{3})}}{2}$

$=4.622$ or $-0.158$