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

1 Answer
Feb 22, 2018

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