# SOLVE to the nearest four decimals?

## $$  $\sqrt{3} \cdot {\left(\cot x\right)}^{2} + \cot x = 1$

##### 2 Answers
May 17, 2018

$x = {62}^{\circ} 3393 + k {180}^{\circ}$
$3 x = - {42}^{\circ} 2363 + k {180}^{\circ}$

#### Explanation:

Method 1
$\sqrt{3} \left({\cos}^{2} \frac{x}{{\sin}^{2} x}\right) + \cos \frac{x}{\sin x} = 1$
$\sqrt{3.} {\cos}^{2} x + \sin x . \cos x = {\sin}^{2} x$
Divide both side by cos x, (condition cos x != 0)
$\sqrt{3} + \tan x = {\tan}^{2} x$
${\tan}^{2} x - \tan x - \sqrt{3} = 0$
Solve this quadratic equation for tan x.
D = d^2 = b^2 - 4ac = 1 + 4sqrt3 = 7.9282 --> $d = \pm 2.8157$
There are 2 real roots:
$\tan x = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{1}{2} \pm \frac{2.8157}{2}$
$\tan x = 0.5 + 1.4079 = 1.9079$
$\tan x = 0.5 - 1.4079 = - 0.9079$
a. $\tan x = 1.9079$
Calculator and unit circle give:
$x = {62}^{\circ} 3393 + k {180}^{\circ}$
b. $\tan x = - 0.9079$
$x = - {42}^{\circ} 2363 + k {180}^{\circ}$

May 17, 2018

$x = {62}^{\circ} 3405 + k {180}^{\circ}$
$x = - {42}^{\circ} 2614 + k {180}^{\circ}$

#### Explanation:

Method 2
Call cot x = t, we get a quadratic equation to solve:
$\sqrt{3} {t}^{2} + t - 1 = 0$
$D = {d}^{2} = {b}^{2} - 4 a c = 1 + 4 \sqrt{3} = 7.9282$ --> $d = \pm 2.8157$
There are 2 real roots:
$t = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{1}{2 \sqrt{3}} \pm \frac{2.8157}{2 \sqrt{3}} =$
$= - 0.2887 \pm 0.8128$
cot x = t = 0.5241 --> $\tan x = 1.9080$
cot x = t = - 1.1015 --> $\tan x = - 0.9078$
a. tan x = 1.9080
$x = {62}^{\circ} 3405 + k {180}^{\circ}$
b. tan x = - 0.9087
$3 x = - {42}^{\circ} 2614 + k {180}^{\circ}$