# Thread: Is there a systematic way to solve trig equations for all zero crossings?

1. ## Is there a systematic way to solve trig equations for all zero crossings?

I'm taking far too long to solve these trig equations. I can't seem to find anything that gives me a systematic way to finding all solutions to a trig equation.

For example, I have $\displaystyle f(x) = \sin (2\pi x)$

It's not particularly difficult differentiate and solve this for x, and end up with:

$\displaystyle x = \frac {\cos^{-1}(0)}{2\pi} = \frac {1}{4}$

I can figure that another zero crossing occurs at $\displaystyle -\frac {1}{4}$, and observe that it is zero every $\displaystyle \frac {1}{2}$. But I find myself lacking a more systematic way of approaching these kinds of problems and solving them more efficiently that I do, currently.

Another short example of one I did yesterday. $\displaystyle f(x) = x + 2 \sin x \implies f'(x) = cos (2\pi x) \implies x = \cos^{-1}(-\frac {1}{2})$. After much bashing of head against wall, I worked out that the solutions are all described by $\displaystyle (1 + 2n)\pi \pm \frac {\pi}{3}~ for ~n \in \mathbb{Z}$. But I can't help but think there's a better way to figure this stuff out. It took a lot of head scratching to figure that one out exactly.

So is there some way I can more easily and quickly find every point at which trig equations make a zero crossing, no matter the period and such?

Hope so, because they are becoming a bit of a pain.

Any help or pointers to this information extremely appreciated, and would save me much time and annoyance.

2. Hello Grep

I think what you need is to study the following equations. They do provide general solutions to the sort of problems you've identified:

For all integer values of $\displaystyle n$:

• $\displaystyle \sin A = \sin B \Rightarrow A = n\pi + (-1)^nB$

• $\displaystyle \cos A = \cos B \Rightarrow A = 2n\pi \pm B$

Check them out for various values of $\displaystyle n$.

The first one gives:

$\displaystyle n = 0: A = B$

$\displaystyle n = 1: A = \pi - B$

$\displaystyle n = 2: A = 2\pi + B$

$\displaystyle n = -1: A = -\pi - B$

... and so on.

The second one is easier:

$\displaystyle n = 0: A = \pm B$

$\displaystyle n = 1: A = 2\pi \pm B$

$\displaystyle n = -1: A = -2\pi \pm B$

... and so on.

Does that help?

3. It does help, sort of. The knowledge that there's a general solution makes me very happy.

Just trying this out a bit, and not 100% sure I get it. But I'm actually really tired right now. Not the best time to figure things out... lol

Let me see if I can apply this. Say, with this equation: $\displaystyle f'(x) = \cos (2\pi x)$.

Ok, so I have $\displaystyle \cos A = \cos B \implies A = 2n \pi \pm B$.

So if I get this right, I solve for B and get $\displaystyle \frac {1}{4}$. So
$\displaystyle n = 0: A = 2(0)\pi \pm \frac {1}{4} = \pm \frac {1}{4}$

$\displaystyle n = 1: A = 2(1)\pi \pm \frac {1}{4} = 2\pi \pm \frac {1}{4}$
etc...
But I also know that $\displaystyle \pm \frac {3}{4}$ is a solution, and I don't see that. So still a bit confused.

Am I close, here? lol I really should get some sleep, then tackle this, but it's bugging me.

Thanks a million for helping me, it's always very appreciated!

Edit: Yeah, I really don't think I got it right. For one, B should be $\displaystyle 2\pi x$, I think.

4. Eureka, I think I've got it.

So $\displaystyle cos (2\pi x) = cos (2\pi \frac {1}{4}) = cos (\frac {\pi}{2})$. And $\displaystyle \frac {\pi}{2}$ is my angle B.

Thus,
$\displaystyle n = 0: x = \pm \frac {\frac {\pi}{2}}{2\pi} = \pm \frac {1}{4}$

$\displaystyle n = 1: x = \frac {2\pi - \frac {\pi}{2}}{2\pi} = \frac {3}{4}$ and $\displaystyle x = \frac {2\pi + \frac {\pi}{2}}{2\pi} = \frac {5}{4}$

Having to divide by $\displaystyle 2\pi$ because my angle B was $\displaystyle 2\pi x$ not just x, and I want to isolate the x.

Thankyouthankyouthankyou!

Now I can sleep... lol

5. Excellent. You've got it!