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Trig problem I can't see into

  • Thread starter Geekchick
  • Start date
  • #1
74
0

Homework Statement



[tex]\sqrt{}3[/tex]sinx-3cosx=0 solve in interval [0,2[tex]\pi[/tex])

Homework Equations



reciprocal identities, pythagorean identities, confunction identities, Even/odd identities, sum/difference formulas, double angle formulas, power reducing formulas, half-angle 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.

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!
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31
[tex]\sqrt{3}sinx-3cosx=0 \Rightarrow \sqrt{3}sinx=3cosx[/tex]

Can you divide by cosx and go from there?
 
  • #4
74
0
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
dynamicsolo
Homework Helper
1,648
4
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/Trigonometric_identity#Linear_combinations

EDIT: Damn, too late and a longer method. lol
The "auxiliary angle" method is fine -- it's just "overkill" on the equation

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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