How do you solve $cos x = cot x$ over the interval 0 to 2pi?

Question

6013 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-01T07:45:18

It is

$cos x = cot x=>cosx=(cosx)/sinx=$

First it should be (so the denominator can be defined)

enter image source here

Hence we have that

$cos x = cot x=>cosx=(cosx)/sinx=>cosx*sinx=cosx=> cosx*(sinx-1)=0=> cosx=0 or sinx=1$

From $cosx=0=>x=pi/2 or x=3pi/2$

From $sinx=1=>x=pi/2$

Hence the solutions are $x=pi/2,x=3pi/2$

How do you solve cos x = cot xcos x = cot x over the interval 0 to 2pi?