
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.
$\displaystyle
3sin(2\theta)=cos(2\theta)
$
This is how I have tried to solve it using Trig identities.
$\displaystyle
3(2sin\theta cos\theta)=sin^2\thetacos^2\theta
$
$\displaystyle
6sin\theta cos\theta(sin^2\theta+cos^2\theta)=0
$
$\displaystyle
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

My mistake, I believe that $\displaystyle sin^2\thetacos^2\theta$ should be $\displaystyle cos^2\thetasin^2\theta$.
It is still getting me into the same place though.

Assuming that values of $\displaystyle \displaystyle \theta$ that make $\displaystyle \displaystyle \cos{2\theta} = 0$ are not possible solutions (I'll leave you to check this)
$\displaystyle \displaystyle \frac{3\sin{2\theta}}{\cos{2\theta}} = 1$
$\displaystyle \displaystyle 3\tan{2\theta} = 1$
$\displaystyle \displaystyle \tan{2\theta} = \frac{1}{3}$.
Go from here...

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

Wow, thank you both for the very quick answer. Looking at that now it seems obvious, however I have one small hangup 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,
$\displaystyle
3tan2\theta=1
$
$\displaystyle
tan2\theta=\frac{1}{3}
$
$\displaystyle
2\theta = 18.43...
$
$\displaystyle
\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?

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

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.
$\displaystyle
tan2\theta= \frac{1}{3}
$
$\displaystyle
2\theta = tan^^1(\frac{1}{3}) = \approx.322
$
$\displaystyle
\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 longabout fashion. Which you can see if you look at my code.

You actually do need the $\displaystyle \displaystyle + 2\pi n$, because if
$\displaystyle \displaystyle 2\theta = \left\{\arctan{\frac{1}{3}}, \pi + \arctan{\frac{1}{3}}\right\} + 2\pi n$ then
$\displaystyle \displaystyle \theta = \left\{\frac{1}{2}\arctan{\frac{1}{3}}, \frac{\pi}{2} + \frac{1}{2}\arctan{\frac{1}{3}}\right\} + \pi n$.
It's quite possible that since the period is diminished, then there are more solutions in the interval.

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.