Question

How do I solve for 0 ≤ x < 2π using the equation cot² x - csc x =1 ??

Answer 1

`x = pi/4, 3/4 pi and 3/2 pi`

Explanation:

`cot^2 x - csc x =1`

`cos^2 x /sin^2 x - 1/sin x =1`

`cos^2 x /sin^2 x - sin x/sin^2 x =1`

`(cos^2 x- sin x)/sin^2 x =1`

`cos^2 x- sin x=sin^2 x`

change `cos^2 x to 1 -sin^2 x`
`1 -sin^2 x -sin x = sin ^2 x`

rearrange the equation
`0 = 2 sin^2 x + sin x -1`

factorize the equation
`0 = (2 sin x -1)(sin x + 1)`

therefore,
`2 sin x -1 =0 and sin x + 1 =0`
#sin x = 1/2 and sin x = -1

`x = pi/4, 3/4 pi` and `x =3/2 pi`