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Help with Trigonometry equation

  1. Jul 31, 2007 #1
    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?
  2. jcsd
  3. Jul 31, 2007 #2
    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.
  4. Aug 1, 2007 #3
    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 dont understand why the k belongs on the end, because I already solved for K, it seems like Im using intuition rather than a logical algebraic method for why it goes back on the end.
  5. Aug 1, 2007 #4
    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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