Question

How do you solve `sqrt3cscx-2=0`?

Answer 1

`{x | x = pi/3 + 2kpi or x = (2pi)/3 + 2kpi, k in ZZ}`

Explanation:

To solve this equation, we need to isolate `x`, which means isolating the trigonometric function that has `x` in it.
If you find it hard to remember the unit circle values for secant or cosecant, you can always use the more familiar sine and cosine functions:

`sqrt(3)csc(x) - 2 = 0`

`sqrt(3)*1/sin(x) - 2 = 0`

`sqrt(3)*1/sin(x) = 2`

`sqrt(3) = 2sin(x)`

`sqrt(3)/2 = sin(x)`

The values for `x` where `sqrt(3)/2 = sin(x)` are `pi/3` and `(2pi)/3` radians.

Since there's no specified domain in the problem, the answer is the general solution.
The `sin` function has the period `2pi`, `pi/3` and `(2pi)/3` repeat every `2pi` in the answer.

`{x | x = pi/3 + 2kpi or x = (2pi)/3 + 2kpi, k in ZZ}`