# Math Help - Trigonometric Equation

1. ## Trigonometric Equation

$0^o \leq x \leq 360^o$

Solve this equation in terms of the angle of x
$
cot^2x-\sqrt{3}cotx=0
$

2. Hello,

Let y be cot x.

The problem is now to solve $y^2 - \sqrt{3} y = 0$

-> $y=0 \ \text{or} \ y=\sqrt{3}$

Between 0° and 180° the cot function is decreasing. So there is one and only one value verifying y=0 in this interval. And the function is pi(180)-periodic.

cot x = 0 when x=90° or 270° (because cot x = cos(x)/sin(x))

We know that $\cos(30)=\sqrt{3}/2$ and $\sin(30)=1/2$
-> $cot(30)=\sqrt{3}$

By the property of the periodicity, $cot(210)=cot(210)=\sqrt{3}$ too

Solutions are $S=\{0;30;90;210;270\}$