Trig Equation

**Sleight** · March 12th 2011, 08:09 PM

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

**Sleight** · March 12th 2011, 08:12 PM

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.

**Prove It** · March 12th 2011, 08:20 PM

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...

**islam** · March 12th 2011, 08:21 PM

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..

**Sleight** · March 12th 2011, 08:33 PM

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?

**Prove It** · March 12th 2011, 08:38 PM

There will be more than one answer...

Assuming you have used radians...

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.

**Sleight** · March 12th 2011, 08:56 PM

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.

**Prove It** · March 12th 2011, 09:01 PM

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.

**Sleight** · March 12th 2011, 10:20 PM

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.

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