1. ## Trig Equation

Hello everyone, this is my first post here and I am very glad to have discovered this forum. I will try very hard to get the Latex correct, but please forgive me if I mess something up.

I am very stuck on the following question.

$
3sin(2\theta)=cos(2\theta)
$

This is how I have tried to solve it using Trig identities.

$
3(2sin\theta cos\theta)=sin^2\theta-cos^2\theta
$

$
6sin\theta cos\theta-(sin^2\theta+cos^2\theta)=0
$

$
sin\theta(6cos\theta)-1=0
$

I have a hunch that I should just stop here, because I can tell I've done something wrong. I would continue on trying to find sin=1 and cos=1, but it just doesn't work.

Any help that can be offered would be greatly appreciated. Also, \\ isn't giving me a new line in latex. Perhaps I am missing something there?

Thanks,
Sleight

2. My mistake, I believe that $sin^2\theta-cos^2\theta$ should be $cos^2\theta-sin^2\theta$.

It is still getting me into the same place though.

3. Assuming that values of $\displaystyle \theta$ that make $\displaystyle \cos{2\theta} = 0$ are not possible solutions (I'll leave you to check this)

$\displaystyle \frac{3\sin{2\theta}}{\cos{2\theta}} = 1$

$\displaystyle 3\tan{2\theta} = 1$

$\displaystyle \tan{2\theta} = \frac{1}{3}$.

Go from here...

hello dear
try to do it like this:
suppose $tanx=3$
find x by using calculator
$tanx=sinx/cosx$
multiply the whole equation by sinx
can u take it from here..

5. Wow, thank you both for the very quick answer. Looking at that now it seems obvious, however I have one small hang-up in the directions of my textbook, and I am not quite sure what it means. It reads like this.
"In algebraic equations, we generally do not divide each side by any expression that involves a variable, and the same rule holds for trigonometric equations."
Because of that sentence (whatever that means) I haven't been cross dividing anything.

So if I follow the workings below,

$
3tan2\theta=1
$

$
tan2\theta=\frac{1}{3}
$

$
2\theta = 18.43...
$

$
\theta = 9.22...
$

Strange though, if that is the correct track. Every other answer has been giving me two answers. I suppose if I shifted that into quadrant 3 I would get 12.36.

Still on the right track?

6. There will be more than one answer...

Since the tangent function is positive in the first and third quadrants, if $\displaystyle \tan{2\theta} = \frac{1}{3}$ then

$\displaystyle 2\theta = \left\{\arctan{\frac{1}{3}}, \pi + \arctan{\frac{1}{3}}\right\} + 2\pi n$, where $\displaystyle n$ is an integer.

7. Yes, I have used radians. One part of this problem that I forgot to mention is that it is in the interval [0,2╥), so I don't need +2k╥. Sorry, I should have mentioned that earlier.

I also noticed that I made an error in my last post. It should look like this.

$
tan2\theta= \frac{1}{3}
$

$
2\theta = tan^-^1(\frac{1}{3}) = \approx.322
$

$
\theta = \approx0.16
$

This works when I plug it back in, so thank you very much. I was not able to get this far on my own. Taking it further, I suppose that I can find the other again within the interval by looking at the unit circle. Doing this I find that the other angle is approximately 3.30..

You guys rock.

Any advice on the // thing for spacing in Latex? I am doing it in sort of a long-about fashion. Which you can see if you look at my code.

8. You actually do need the $\displaystyle + 2\pi n$, because if

$\displaystyle 2\theta = \left\{\arctan{\frac{1}{3}}, \pi + \arctan{\frac{1}{3}}\right\} + 2\pi n$ then

$\displaystyle \theta = \left\{\frac{1}{2}\arctan{\frac{1}{3}}, \frac{\pi}{2} + \frac{1}{2}\arctan{\frac{1}{3}}\right\} + \pi n$.

It's quite possible that since the period is diminished, then there are more solutions in the interval.

9. Your completely right, I had forgotten about the half angle identity involved. Although it appears that these answers are only in quadrant 1 and 3, so they are already ╥ distance away from each other.

Once again, I appreciate all of your help.