Question

cot^2x+cotx=0
The answer has to be between the radians 0 and 2pie

Answer 1

Replace Cotg = y

y^2 + y = 0

Put x as commun factor

y( y + 1) = 0

y = 0

Or

y = -1

Then,

Cotg(x) = 0

Or

Cotg(x) = -1
__________

As Cotg(x) = Cos(x)/Sen(x)

Then,

Cos(x)/Sen(x) = 0

Cos(x) = 0

x = 90° or pi/2


Cos(x)/Sen(x) = -1

Cos(x) = - Sen(x)

As (Senx)^2+ (Cosx)^2= 1

Then,

(Senx)^2 + (-Sen x)^2 = 1

2.(Senx)^2 = 1

(Senx)^2 = 1/2

Sen x = √1 / √2

Sen x = √2 / 2

x = pi/4

But As Cos(x) = - Sen(x)

x is on the 2 quadrant

x = pi/4 + pi

x = pi/4 + 4pi/4

x = 5pi/4

Cos(5pi/4) = - Sen(5pi/4)

It is correct, do the test.

5pi/4 = 225°