$\displaystyle 0^o \leq x \leq 360^o$
Solve this equation in terms of the angle of x
$\displaystyle
cot^2x-\sqrt{3}cotx=0
$
Hello,
Let y be cot x.
The problem is now to solve $\displaystyle y^2 - \sqrt{3} y = 0$
-> $\displaystyle y=0 \ \text{or} \ y=\sqrt{3}$
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 $\displaystyle \cos(30)=\sqrt{3}/2$ and $\displaystyle \sin(30)=1/2$
-> $\displaystyle cot(30)=\sqrt{3}$
By the property of the periodicity, $\displaystyle cot(210)=cot(210)=\sqrt{3}$ too
Solutions are $\displaystyle S=\{0;30;90;210;270\}$