
#1
Oct605, 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 



#2
Oct605, 03:46 PM

HW Helper
P: 930

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
Oct605, 04:22 PM

HW Helper
P: 1,024





#4
Oct605, 04:34 PM

P: 786

Another Trig EquationDoes 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
Oct605, 04:38 PM

HW Helper
P: 1,024

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] 



#6
Oct605, 04:48 PM

P: 786

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. 



#7
Oct605, 04:59 PM

HW Helper
P: 1,024

It will if you use an array:
[tex]\begin{array}{l} x^2  4 = 0 \\ x = 2 \vee x =  2 \\ \end{array}[/tex]




#8
Oct605, 05:39 PM

P: 786

Doh! Thanks again!




#9
Oct605, 05:39 PM

HW Helper
P: 1,024

You're welcome



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