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Trig Problem- solve equation. Check

  1. Jan 16, 2012 #1
    1. The problem statement, all variables and given/known data

    Solve the equation 5sin2x-3=0, given values of x in the interval -180≤x≤180 correct to 1 decimal place.

    2. Relevant equations
    for sin: take inverse
    180-θ
    ±360


    3. The attempt at a solution
    I rearranged the equation to
    sin2=3/5
    but Im not quite sure what to do since sine is squared. I think it wont affect answer.

    Then I did the inverse which was 36.9
    subtrated 180 from inverse; 180- 36.9 = 143.1
    Then I did 360±36.9 and 360±143.1
    where both of these didn't fit in the interval.

    Is this correct: x=36.9° and 143.1°
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 16, 2012 #2

    SammyS

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    What do you mean by " ... I'm not quite sure what to do since sine is squared. I think it wont affect answer." ?

    Is the solution to u = 5/3 the same as the solution to u2 = 5/3 ? ... Of course not.
     
  4. Jan 16, 2012 #3
    Aww. Alright. Then how would I do it if sin is squared? Sin^2=3/5
     
  5. Jan 16, 2012 #4

    SammyS

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    That should be sin2(x) = 3/5.

    Take the square root of both sides of the above equation just like you would to solve u2 = 3/5 .

    Don't forget the ± .

    After that use the arc-sine function, sin-1 .

    Alternatively, you can use the identity, cos(2x) = 1 - 2sin2(x).
     
  6. Jan 17, 2012 #5
    Ohhh. So it would be sinx= +/- the sqr root 3/5. Which then the inverse would be 50.8. Then 180 minus 50.8 is 129.2. After that I would take those two answers and +/- 360 which doesn't work. So my two answers I got are 50.8 and 129.2.
     
  7. Jan 17, 2012 #6

    SammyS

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    Those two answers are fine. They're for [itex]\displaystyle \sin(x)=\sqrt{\frac{3}{5}}\,.[/itex]

    Now find the solution for the ' - ' sign: [itex]\displaystyle \sin(x)=-\sqrt{\frac{3}{5}}\,.[/itex]
     
  8. Jan 17, 2012 #7
    Okay, for negative square root:
    The inverse is -50.8
    180-(-50.8). Is 230.8 which doesn't fit the range
    But if I subtract from 360 I get -129.2
    So my my two answers for this is -50.8 and -129.2 giving me a total of four answers
     
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