Combining like terms

Question

Sinx-cot^2x=1 

0<x<2pi

 

I'm having trouble with this problem.

Answer 1

sinx - cot²x = 1

sinx - cos²x / sin²x = 1

(sin³x - cos²x) / sin²x = 1

sin³x - cos²x = sin²x

sin³x = sin²x + cos²x

sin³x = 1

Take cube root of both sides:

sin x = 1

arcsin 1 = 1.57 radians OR pi/2

Answer 2

sinx - cot²x = 1

 

sinx - cos²x/sin²x = 1

 

sinx - cos²x/sin²x - 1 = 0

 

(sin³x - cos²x - sin²x)/sin²x = 0

 

[sin³x - (cos²x + sin²x)]/sin²x = 0

 

(sin³x - 1)/sin²x = 0

 

Set the numerator = 0

 

sin³x - 1 = 0

 

sin³x = 1

 

sinx = 1       x = π/2