Question

How do you solve `sec^2x-2tan^2x=0`?

Answer 1

`x=pi/4+(kpi)/2,kinZZ`

Explanation:

Here, we can use the Pythagorean Identity, which states that `1+tan^2x=sec^2x`.

We can substitute this into the original equation to write it just in terms of tangent, instead of with both tangent and secant.

`(sec^2x)-2tan^2x=0" "=>" "(1+tan^2x)-2tan^2x=0`

Simplifying this by combining the `tan^2x` terms gives:

`1-tan^2x=0`

Thus:

`tan^2x=1`

Taking the square root, and remembering both the positive and negative roots:

`tanx=+-1`

Notice that `tanx=1` at `x=pi/4` and `x=(5pi)/4`, and all angles with a reference angle of `pi/4` in quadrants `"I"` and `"III"`.

Similarly, `tanx=-1` at `x=(3pi)/4` and `(7pi)/4` and all other angles with a reference angle of `pi/4` in quadrants `"II"` and `"IV"`.

Combining all these answers, and remembering that they go on forever, we see that `x=pi/4,(3pi)/4,(5pi)/4,(7pi)/4,(9pi)/4`, with `x=(9pi)/4` being exactly one `2pi` revolution away from `pi/4`.

But, the solutions don't stop, and we can generalize them by making a rule. Note that every solution for `x` is `(2pi)/4=pi/2` away from the next solution. So, we can think about starting at `x=pi/4` then adding some whole number amount of `pi/2` values to the starting value.

This is expressed as

`x=pi/4+(kpi)/2,kinZZ`

Note that `kinZZ` just means that `k` is an integer, which includes numbers like `1,-5,7,` and `-68`.