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Trig problem I can't see intoby Geekchick
Tags: trig 
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#1
Aug1608, 09:43 PM

P: 74

1. The problem statement, all variables and given/known data
[tex]\sqrt{}3[/tex]sinx3cosx=0 solve in interval [0,2[tex]\pi[/tex]) 2. Relevant equations reciprocal identities, pythagorean identities, confunction identities, Even/odd identities, sum/difference formulas, double angle formulas, power reducing formulas, halfangle formulas, sum to product formulas, product to sum formulas. I am not sure how many of those are relevant but thats all the identities and formulas I have learned so far. 3. The attempt at a solution To be quite honest I don't even know where to start with this problem : ( mostly the square root of three is throwing me. I have been trying to solve this for days so if anyone could so much as tell me what formulas or identites to use i would appreciate it. I'm just so lost. Oh also I'm new to this forum, just found it tonight, so i just want to say hi to everyone! 


#2
Aug1608, 09:58 PM

HW Helper
P: 6,207

[tex]\sqrt{3}sinx3cosx=0 \Rightarrow \sqrt{3}sinx=3cosx[/tex]
Can you divide by cosx and go from there? 


#3
Aug1608, 10:01 PM

HW Helper
P: 3,352

Welcome to PF Geekchick =]
In general, for things like a sin x +/ b cos x, we can combine them into a single sine term, using something known as the Auxiliary Angle method: http://en.wikipedia.org/wiki/Trigono...r_combinations EDIT: Damn, too late and a longer method. lol 


#4
Aug1608, 10:16 PM

P: 74

Trig problem I can't see into
Thank you! Thank you! thank you! I didn't even think about the fact that sin/cos is equal to tan duh! *smacks head* I was trying to make things way more complicated. So just to be sure once I divide by cos I'm left with the (square root of 3)tan=3 which after dividing that by the (square root of 3) and simplifying I have tan= square root of 3 which in the interval [0,2pi) means my answer is (pi/3) and (4pi/3). Yay!



#5
Aug1908, 12:15 AM

HW Helper
P: 1,663

a sin x + b cos x = c in the case where c = 0. When you follow through with it for this problem, you get [tex]sin(x  \frac{4 \pi}{3}) = 0 \Rightarrow x  \frac{4 \pi}{3} = ..., \pi, 0, \pi, ...[/tex] with the results in the fundamental circle being the two Geekchick has found. (I'm just elaborating on this since it's always good to know multiple methods for solving a problem and to confirm that they all give the same answer.) 


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