Question

Solve the following inequality:
${\tan }^{2}x-(1+\sqrt{3})\tan x+\sqrt{3}>0$

Answer 1

Solution-

$ta{n}^{2}x-(1+\sqrt{3})tanx+\sqrt{3}>0$

$ta{n}^{2}x-tanx-\sqrt{3}tanx+\sqrt{3}>0$

$tanx\left(1tanx\right)+\sqrt{3}(-tanx+1)>0$

$(tanx+\sqrt{3})(1-tanx)>0$

$(\sqrt{3}+tanx)(1-tanx)>0$

$\downarrow$ $\downarrow$

$X$ $Y$

XY is positive when both X & Y are positive or negative

Lets both X,Y are positive.

$tanx>-\sqrt{3}$ and $tanx<1$

$xϵ((0,\frac{\pi }{2})\cup (\frac{2\pi }{3},\pi ))\cap \left((0,\frac{\pi }{4}\cup (\frac{\pi }{2},\pi ))\right)$

$xϵ(0,\frac{\pi }{4})\cup (\frac{2\pi }{3},\pi )$

Lets both X, Y are negative.

$tanx<-\sqrt{3}.$ and $tanx>1$

$x\epsilon (\frac{\pi }{2},\frac{2\pi }{3})\cap (\frac{\pi }{4},\frac{\pi }{2})=\varphi$

General solution-

$xϵ(x\pi ,x\pi +\frac{\pi }{4})\cup (x\pi +\frac{2\pi }{3},x\pi +\pi )=x=$ integer