Solving Trig Problem with Square Root: [0,2π) Interval | Forum Help Needed

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The problem involves solving the equation √3sin(x) - 3cos(x) = 0 within the interval [0, 2π). The initial approach suggests dividing by cos(x) to simplify the equation to √3tan(x) = 3, leading to tan(x) = √3. The solutions for x in the specified interval are found to be π/3 and 4π/3. The discussion also highlights the auxiliary angle method as an alternative approach, confirming that multiple methods yield the same results. Overall, the thread provides guidance on tackling trigonometric equations effectively.
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Homework Statement



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

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 that's 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!
 
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\sqrt{3}sinx-3cosx=0 \Rightarrow \sqrt{3}sinx=3cosx

Can you divide by cosx and go from there?
 
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!
 
Gib Z said:
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

sin(x - \frac{4 \pi}{3}) = 0 \Rightarrow x - \frac{4 \pi}{3} = ..., -\pi, 0, \pi, ...

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.)
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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