1. The problem statement, all variables and given/known data [tex]\sqrt{}3[/tex]sinx-3cosx=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, 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. 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!
[tex]\sqrt{3}sinx-3cosx=0 \Rightarrow \sqrt{3}sinx=3cosx[/tex] Can you divide by cosx and go from there?
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/Trigonometric_identity#Linear_combinations EDIT: Damn, too late and a longer method. lol
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!
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.)