## Another Trig Equation

\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

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 Recognitions: Homework Help 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.

Recognitions:
Homework Help
 Quote by cscott \begin{align*} \sin \theta = \cos \theta\\ \theta = \frac{\pi}{4} + 2\pi k, k \epsilon \mathbb{I} \end{align*}
Here, you can get all solution by taking pi/4 + k*pi instead of 2k*pi.

## Another Trig Equation

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

 Recognitions: Homework Help How's that? $$\sin ^2 \theta + 2\sin \theta \cos \theta - 3\cos ^2 \theta \mathop \to \limits^{\theta = \pi /4} \left( {\sin \frac{\pi }{4}} \right)^2 + 2\sin \frac{\pi }{4}\cos \frac{\pi }{4} - 3\left( {\cos \frac{\pi }{4}} \right)^2 = \frac{1}{2} + 2 \cdot \frac{{\sqrt 2 }}{2} \cdot \frac{{\sqrt 2 }}{2} - 3 \cdot \frac{1}{2} = 0$$

 Quote by TD How's that? $$\sin ^2 \theta + 2\sin \theta \cos \theta - 3\cos ^2 \theta \mathop \to \limits^{\theta = \pi /4} \left( {\sin \frac{\pi }{4}} \right)^2 + 2\sin \frac{\pi }{4}\cos \frac{\pi }{4} - 3\left( {\cos \frac{\pi }{4}} \right)^2 = \frac{1}{2} + 2 \cdot \frac{{\sqrt 2 }}{2} \cdot \frac{{\sqrt 2 }}{2} - 3 \cdot \frac{1}{2} = 0$$
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.

 Recognitions: Homework Help It will if you use an array: $$\begin{array}{l} x^2 - 4 = 0 \\ x = 2 \vee x = - 2 \\ \end{array}$$ Code: \begin{array}{l} x^2 - 4 = 0 \\ x = 2 \vee x = - 2 \\ \end{array}
 Doh! Thanks again!