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

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Homework Help Overview

The problem involves solving the equation \(\sqrt{3}\sin x - 3\cos x = 0\) within the interval \([0, 2\pi)\). The context is trigonometric identities and equations, particularly focusing on the relationship between sine and cosine functions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster expresses uncertainty about how to start solving the equation, particularly regarding the square root of three. Some participants suggest dividing by cosine to simplify the equation, while others mention using the Auxiliary Angle method to combine sine and cosine terms.

Discussion Status

Participants are exploring different methods to approach the problem. Some guidance has been offered regarding the use of trigonometric identities and simplification techniques, but there is no explicit consensus on a single method being preferred.

Contextual Notes

The original poster is new to the forum and has expressed feeling lost with the problem. There is a mention of various trigonometric identities learned, but uncertainty remains about their relevance to the current problem.

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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.)
 

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