# Another Trig Equation

1. Oct 6, 2005

### cscott

\begin{align*} \sin^2 \theta + 2 \sin \theta \cos \theta - 3 \cos^2 \theta = 0 \\ (\sin \theta + 3 \cos \theta)(\sin \theta - \cos \theta) = 0 \\ \end{align*}

so,

\begin{align*} \sin \theta = \cos \theta\\ \theta = \frac{\pi}{4} + \pi k, k \epsilon \mathbb{I} \end{align*}

or

\begin{align*} \sin \theta = -3 \cos \theta\\ \tan \theta = -3\\ \theta = 5.03 + 2\pi k, k \epsilon \mathbb{I}; 1.89 + 2\pi k, k \epsilon \mathbb{I} \end{align*}

How come this is incorrect?

...argh I can't align it properly :uhh:

Last edited: Oct 6, 2005
2. Oct 6, 2005

### hotvette

a x b = 0 doesn't mean both a and b are zero, although they could be. It simply means at least one of them is zero for the product to be zero.

3. Oct 6, 2005

### TD

Here, you can get all solution by taking pi/4 + k*pi instead of 2k*pi.

4. Oct 6, 2005

### cscott

Oops, I had that written down but I made the mistake when posting!

Does this mean you say pi/4 is a correct answer? When I plug it back into the original equation I don't get 0

5. Oct 6, 2005

### TD

How's that?

$$\sin ^2 \theta + 2\sin \theta \cos \theta - 3\cos ^2 \theta \mathop \to \limits^{\theta = \pi /4} \left( {\sin \frac{\pi }{4}} \right)^2 + 2\sin \frac{\pi }{4}\cos \frac{\pi }{4} - 3\left( {\cos \frac{\pi }{4}} \right)^2 = \frac{1}{2} + 2 \cdot \frac{{\sqrt 2 }}{2} \cdot \frac{{\sqrt 2 }}{2} - 3 \cdot \frac{1}{2} = 0$$

6. Oct 6, 2005

### cscott

OK, I'm just bad at typing things into my calculator then! That's what I get for using it in the first place

By the way, can anyone tell me how to align things properly with tex? For some reason using "\\" won't skip lines for me.

Thanks.

Last edited: Oct 6, 2005
7. Oct 6, 2005

### TD

It will if you use an array:

$$\begin{array}{l} x^2 - 4 = 0 \\ x = 2 \vee x = - 2 \\ \end{array}$$

Code (Text):

\begin{array}{l}
x^2  - 4 = 0 \\
x = 2 \vee x =  - 2 \\
\end{array}

8. Oct 6, 2005

### cscott

Doh! Thanks again!

9. Oct 6, 2005

### TD

You're welcome