Help with Trigonometry equation

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Discussion Overview

The discussion revolves around the algebraic manipulation of trigonometric equations, specifically how to derive a general solution for the equation 6 sin²(X) - 3 = 0. Participants explore the transition from one form of the general solution, X = (pi)/4 + 2(pi)K, to another form, X = (pi)/4 + (pi)/2K, while addressing the completeness of the solutions.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant asks for the algebraic method to transform X = (pi)/4 + 2(pi)K into X = (pi)/4 + (pi)/2K.
  • Another participant points out that the initial solution does not account for all values of X that satisfy sin(X) = +/- 1/sqrt(2), suggesting that additional solutions exist within the interval [0, 2pi).
  • A different participant proposes that since all solutions are multiples of pi/4, the reasoning for including K in the final form is based on periodicity, as adding multiples of (pi/2) generates all solutions.
  • One participant expresses confusion about the necessity of K in the final equation, having solved for a specific value of K but questioning its relevance in the general solution.
  • Another participant clarifies that while the equation X = (pi)/4 + 2(pi)K is correct, it does not encompass all solutions due to K being restricted to integers, leading to the conclusion that X = (pi)/4 + (pi)/2K is a more comprehensive representation.

Areas of Agreement / Disagreement

Participants express differing views on the completeness of the initial solution and the necessity of K in the final equation. There is no consensus on whether the transformation can be achieved algebraically or if it is purely a matter of understanding the periodic nature of the sine function.

Contextual Notes

Participants highlight the dependence on the periodicity of the sine function and the implications of integer versus fractional values for K in the context of generating all solutions. The discussion remains focused on the algebraic manipulation and reasoning rather than definitive conclusions.

popsquare
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What is the algebraic method to get from X = (pi)/4 + 2(pi)K into this form => X = (pi)/4 + (pi)/2K ? These are the general equations for all solutions to this original equation: 6 sin^2 X-3=0 , I can understand how to get the general equation graphically can someone show me how to get from the first general equation to the second general equation which is the correct version?
 
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when you solve 6*sin^2(x)-3=0, you get sin(x)=+/- 1/sqrt(2). So, the first answer you gave is x=(pi/4)+2*pi*K. But this isn't all the answers. There are other such x values in [0,2pi) that satisfy sin(x)=+/- 1/sqrt(2) and they are (3pi)/4, (5pi)/4 and (7pi)/4. So you need to modify your solution to start with pi/4 and add these other values and all their multiples, which basically comes down to your second solution x=(pi/4)+(pi/2)*K since (3pi)/4=(pi/4)+(pi/2)*1, (5pi)/4=(pi/4)+(pi/2)*2 and so on.
 
i guess the solution needs to be modified since they are all multiples of pi/4. There might be a better way to understand this , I solved
3*pi/4 = pi/4 + 2*pi*k for k and got k=1/4. Then plugged this into the general equation for all solutions and got, pi/4+2*pi*1/4 which becomes
pi/4+pi/2 and just stuck a k on the end to get the general solution of
pi/4+pi/2*k. Is this correct reasoning because I don't understand why the k belongs on the end, because I already solved for K, it seems like I am using intuition rather than a logical algebraic method for why it goes back on the end.
 
in your last post, you are showing algebraically what K needs to be in order to get (3pi)/4. But the K must stay in the equation because sin is a periodic function. Like I said, pi/4, (3pi)/4, (5pi)/4 and (7pi)/4 all work as solutions. But if you add 2pi to any of these they always work. So the equation that accounts for all of this is x=(pi/4)+(pi/2)*K, where K is any integer, since to get from one solution to the next you just have to add some multiple of (pi/2).

so to sum it all up: the equation x=(pi/4)+2*pi*K is a correct equation but if K can only be an integer, it doesn't get all of the solutions. Like you showed, in order to get (3pi)/4 you need K=1/4. So rather than keeping that equation where this K can be fractions, they changed the equation to x=(pi/4)+(pi/2)*K, where K is any integer, which gives all the solutions in the most simple manner. There is not a way to algebraically manipulate the first answer to get the second one. It is just the reasoning of what the solution actually is and what the best way to present it would be.
 

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