Another Trig Equation

[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:

[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.

[TD]

\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.

[cscott]

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 :frown:

[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

[cscott]

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 o:)

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.

[TD]

It will if you use an array:

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


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

[cscott]

Doh! Thanks again!

[TD]

You're welcome :smile: