Question

How do you find the general solutions for `cot^2 x +csc x = 1`?

Answer 1

`x = n pi - pi/6(-1)^n " or " x = 2npi + pi/2 ", " n in ZZ`

Explanation:

`sin^2 x + cos^2 x = 1`
Divide by `sin^2 x` gives: `1 + cot^2 x = csc^2 x`

Therefore: `csc^2 x + csc x - 2 = 0`
`(csc x + 2)(csc x - 1) = 0`
`1/sin x = -2 " or " 1/sin x = 1`
`sin x = -1/2 " or " sin x = 1`
`x = n pi - pi/6(-1)^n " or " x = 2npi + pi/2 ", " n in ZZ`