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Solution of Trigonometric Equations

  1. Aug 31, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the solution set of each of the following equations for the interval 0° ≤ x ≤ 360°.
    [tex]sin^2x = sinx[/tex]

    2. Relevant equations
    Trigonometric Identities for Sine.

    3. The attempt at a solution
    This is my attempt so far:

    [tex]\sin^2x = sinx[/tex]
    [tex]\sin^2x - sinx = 0[/tex]
    [tex]\sin^2x - sinx + \frac{1}{4} = \frac{1}{4}[/tex]
    [tex](sinx - \frac{1}{2})^2 = \frac{1}{4}[/tex]
    taking square roots of both sides:
    [tex]sinx -\frac{1}{2} = \frac{1}{2}[/tex]
    [tex]sinx = ±\frac{1}{2} + \frac{1}{2}[/tex]
    if [itex]\frac{1}{2},[/itex]
    [tex]sinx = 1[/tex]
    if [itex]-\frac{1}{2},[/itex]
    [tex]sinx = 0[/tex]
    take arcsin of both sides:
    [tex]x = arcsin1[/tex]
    [tex]x = 90° or \frac{\pi}{2}[/tex]
    [tex]x = arcsin0[/tex]
    [tex]x = 0°[/tex]

    Solution set = {0°, 90°} or {0, [itex]\frac{\pi}{2}[/itex]}

    EDIT: The complete solution set must be:

    x = {0°, 90°, 180°, 360°}
    because sinx = 0 also in 180 and 360, not only in 0°)
    Special thanks to Sourabh N for guiding me to the correct solution o:)
     
    Last edited: Aug 31, 2012
  2. jcsd
  3. Aug 31, 2012 #2
    Notice the range of x: 0° ≤ x ≤ 360°. What other values of x within this range satisfy the conditions x = arcsin1 or x = arcsin0?
     
  4. Aug 31, 2012 #3
    How about this?

    x = {0, 90°, 360°}
     
  5. Aug 31, 2012 #4
    No.

    Here's a very good way to figure out all the solutions. Draw the curve sin[x] for 0° ≤ x ≤ 360°. Also draw horizontal lines at y = 0 and y = 1. The point(s) where sin[x] and these lines intersect, are your solutions.
     
  6. Aug 31, 2012 #5
    Oh. You mean, what values of x does give the angles 0, 90°, 360° when worked with sine?
     
  7. Aug 31, 2012 #6
    for x values,

    x = {0, 1}
     
  8. Aug 31, 2012 #7
    No. x IS your angle.

    y = Sin[x].
    x = x.

    1. Draw y = 0. See at what values of x do Sin[x] and y = 0 intersect.
    2. Same with y = 1.
     
  9. Aug 31, 2012 #8
    Is my graph correct?

    2w3tmc7.png
     
  10. Aug 31, 2012 #9
    On the horizontal axis, you have 0 -> 90 -> 360?? In 0->90-> What comes after 90?
     
  11. Aug 31, 2012 #10
    This?
    2w3wzsl.png
     
  12. Aug 31, 2012 #11
    Very good! You have also (correctly) marked the points of intersection, except that you forgot to include x = 0.

    So what is the solution set?
     
  13. Aug 31, 2012 #12
    Yeah, and also the point at 360. Sorry :shy:

    Based on the graph, the x values are 0°, 45°, 90°, 225°, 360°. Right?
     
  14. Aug 31, 2012 #13
    Ah, I didn't look carefully. Your Sin[x] curve is not correct.

    What is Sin[0]? Sin[45]? Sin[90]? Do you see what's wrong?
     
  15. Aug 31, 2012 #14
    Do you have any clue of the correct curve for sin x? because that's the only sinx graph that I know.
     
  16. Aug 31, 2012 #15
    The shape is correct, but you don't have the scaling right. Sin[x] reaches 1 at x = 90°, Sin[x] reaches back to 0 at 180° and so on.
     
  17. Aug 31, 2012 #16
    Oh yeah I forgot that I have to draw only 1 period. Apologies~ :tongue:
    Working on graph..
     
  18. Aug 31, 2012 #17
    No. You have Sin[x] = 1 at x = 45°, Sin[x] = 0 at x = 90°, which is wrong.

    You have to draw the curve for 0° ≤ x ≤ 360°.
     
  19. Aug 31, 2012 #18
    How about this? I hope this one got it right:
    2hfncb5.png
     
  20. Aug 31, 2012 #19
    That's correct. (Though in he graph, you call y = 0 as y = 2).

    The red dots are your solution set.
     
  21. Aug 31, 2012 #20
    :rofl: that should be this:
    2llzz7t.png

    Sorry about that. I labeled the green line first as y = 1 and accidentally labeled the second one y = 2. What a mess.

    So, the solution set is:

    x = {0°, 90°, 180°, 360°}

    Thank you so much sir! I had fun learning with you. You are good in teaching. I would vote for you as a homework helper here and hope you become one :approve:
     
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