Question

How do I without using my calculator, solve the following equation?

`csc^2`x-4=0

I changed it so that I have `sinx=1/2` and `sinx =-1/2`
The next step I'm a bit unsure of because I got some answers, but half of them were wrong.

Answer 1

`x=pi/6+2npi`
`x=(5pi)/6+2npi`
`x=(7pi)/6+2npi`
`x=(11pi)/6+2npi`
`n` is any integer.

Explanation:

Let's solve the equation for `csc(x):`

`csc^2x-4=0`

`csc^2x=4`

`cscx=+-2`

Knowing that `cscx=1/sinx`,

`1/sinx=+-2`

Flip both sides:

`sinx=+-1/2`

First, let's consider the values of `x` which yield `1/2` for the sine function. This is true for `x=pi/6` which is in the first quadrant. However, it's also true for `x=(5pi)/6,` which is in the second quadrant, as sine is also positive in the second quadrant.

So, `x=pi/6, x=(5pi)/6` give `1/2` for `sinx.`

Now, let's consider which values give us `sinx=-1/2:`

`x=(7pi)/6` in the third quadrant fulfills this; however, so does `x=(11pi)/6.`

So, we have `x=pi/6, (5pi)/6, (7pi)/6, (11pi)/6.` We want to simplify our answer and get an answer for `x` which will cover ALL solutions, even those beyond the interval `[0, 2pi].` Such solutions certainly exist, as the sine function is periodic, meaning that there are infinitely many values for `x` in the interval of all real numbers which give us our desired values for sine.

So, let's look at the following values:
`x=pi/6`
`x=(5pi)/6`
`x=(7pi)/6`
`x=(11pi)/6`

And add `2npi` to all of them, where `n` is any integer. We obtain `2npi` from recognizing that values of sine repeat for every `2pi` units, meaning our answers must repeat every `2pi` units.

`x=pi/6+2npi`
`x=(5pi)/6+2npi`
`x=(7pi)/6+2npi`
`x=(11pi)/6+2npi`