Question

How do you solve the identity `csc^2 x = 3 csc x + 4`?

Answer 1

`S = {arcsin(1/4), pi - arcsin(1/4), (3pi)/2}`

Explanation:

Reewrite `csc(x)` as `y`

`y^2 = 3y + 4`

Make one of the sides equal 0

`y^2 - 3y - 4 = 0`

Solve the quadratic either by sum and product, (wielding `4` and `-1`) or by the formula

`y = (3 +- sqrt(9 -4*1*(-4)))/2 = (3 +- sqrt(9 +16))/2 = (3 +- sqrt(25))/2`
`y= (3 +- 5)/2`
`r_1 = (3+5)/2 = 8/2 = 4`
`r_2 = (3-5)/2 = -2/2 = -1`

So we know that

`csc(x) = 1/sin(x) = -1` or `csc(x) = 1/sin(x) = 4`, which means

`sin(x) = -1` or `sin(x) = 1/4`

So, for answers within the range of `[0,2pi]`, we have that
`sin(x) = -1 rarr x = (3pi)/2`
`sin(x) = 1/4 rarr x = arcsin(1/4) and x = pi - arcsin(1/4)`

So the set of solutions `S` is
`S = {arcsin(1/4), pi - arcsin(1/4), (3pi)/2}`

Or, using approximate values
`S ~= {1/4, (3pi)/4, (3pi)/2}`