Question

How do you solve `csc^2x-2=0` and find all solutions in the interval `[0,2pi)`?

Answer 1

`pi/4; (3pi)/4; (5pi)/4; (7pi)/4`

Explanation:

`1/(sin^2 x) = 2`
`sin^2 x = 1/2`
`sin x = +- 1/(sqrt2) = +- sqrt2/2`
Trig table and unit circle gives 4 solutions:
a. `sin x = sqrt2/2` -->
`x = pi/4` and `x = (3pi)/4`
b. `sin x = - sqrt2/2`
`x = - pi/4`, or `x = (7pi)/4` (co-terminal), and
`x = pi - (-pi/4) = pi + pi/4 = (5pi)/4`