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Another Trig Equation

by cscott
Tags: equation, trig
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cscott
#1
Oct6-05, 03:24 PM
P: 786
[tex]
\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*}
[/tex]

so,

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

or

[tex]
\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*}
[/tex]

How come this is incorrect?

...argh I can't align it properly
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#2
Oct6-05, 03:46 PM
HW Helper
P: 925
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
#3
Oct6-05, 04:22 PM
HW Helper
P: 1,021
Quote Quote by cscott
[tex]\begin{align*}
\sin \theta = \cos \theta\\
\theta = \frac{\pi}{4} + 2\pi k, k \epsilon \mathbb{I}
\end{align*}
[/tex]
Here, you can get all solution by taking pi/4 + k*pi instead of 2k*pi.

cscott
#4
Oct6-05, 04:34 PM
P: 786
Another Trig Equation

Quote 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
TD
#5
Oct6-05, 04:38 PM
HW Helper
P: 1,021
How's that?

[tex]
\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
[/tex]
cscott
#6
Oct6-05, 04:48 PM
P: 786
Quote Quote by TD
How's that?

[tex]
\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
[/tex]
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.
TD
#7
Oct6-05, 04:59 PM
HW Helper
P: 1,021
It will if you use an array:

[tex]\begin{array}{l}
x^2 - 4 = 0 \\
x = 2 \vee x = - 2 \\
\end{array}[/tex]

\begin{array}{l}
 x^2  - 4 = 0 \\ 
 x = 2 \vee x =  - 2 \\ 
 \end{array}
cscott
#8
Oct6-05, 05:39 PM
P: 786
Doh! Thanks again!
TD
#9
Oct6-05, 05:39 PM
HW Helper
P: 1,021
You're welcome


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