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Solving trigonometric equations- URGENT VERY HARD watch

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    Solve the equation 2 sin 3x=1 for -360<x<360
    They got the answers x = −350°, −310°, −230°, −190°, −110°, −70°, 10°, 50°, 130°, 170°, 250°, 290°.
    I will post my working...
    I got half the answers but I don't how they got the rest.
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    -360<x<360, -1080<3x<1080.
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    What do you notice about the negative and the positive numbers? What correlation do they have with each other?
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    Even though the range is from -360 to 360 for x, 3x can be higher than that while still being positive. So they found all the possible positive values for sin3x and then divided them all by 3 after the arcsin, then they fell within this range
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    (Original post by bondangle)
    Even though the range is from -360 to 360 for x, 3x can be higher than that while still being positive. So they found all the possible positive values for sin3x and then divided them all by 3 after the arcsin, then they fell within this range
    (Original post by Y11_Maths)
    What do you notice about the negative and the positive numbers? What correlation do they have with each other?
    (Original post by black1blade)
    -360<x<360, -1080<3x<1080.
    I will post my working in a minute, but I know the range is from -1080<x<1080
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    x= -360 to 360

    2sin(3x) = 1

    Input interpretation:

    Plot[2 Sin[3 x] == 1, {x, -360, 360}]Plot:

    Plot[{2 Sin[3 x], 1}, {x, -360, 360}]
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    the ones I've underlined in red, when divided by 3 they aren't right, I don't know why because i followed the rules of the quadrants properly.
    Same with the minuses, I haven't posted that part put if you guys can tell me where I have gone wrong, then i will apply that to th minus part
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    Compose the result function for -360 < x < 360 by replacing the function designators with the actual function.

    -360 < x < 360

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    (Original post by PsychoAnalyser)
    Compose the result function for -360 < x < 360 by replacing the function designators with the actual function.

    -360 < x < 360

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    Hi, thanks for your reply, however, I am only in year 12 therefore i don't know what that is.
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    anyone
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    Ah, okay...
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    I don't know how to explain this... I was taught this as part of my private tutoring.
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    Please, I have a test on monday Notnek
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    (Original post by Ilovemaths...)
    the ones I've underlined in red, when divided by 3 they aren't right, I don't know why because i followed the rules of the quadrants properly.
    Same with the minuses, I haven't posted that part put if you guys can tell me where I have gone wrong, then i will apply that to th minus part
    Can you please explain how you used the quadrant diagram to get 330?

    sin(330) is not 1/2.
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    isolate the variable-containing term

    Trigonometric Identities might help?

    Pythagorean theorem -THIS WILL HELP! (possibly)
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    (Original post by Notnek)
    Can you please explain how you used the quadrant diagram to get 330?

    sin(330) is not 1/2.
    360-30 because its another revolution you add 180 degrees to 180 making 360 degrees. See in my working out, I do not get how that doesn't give you 1/2?
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    Ok i'm out, sorry, I have to revise.
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    (Original post by PsychoAnalyser)
    isolate the variable-containing term

    Trigonometric Identities might help?

    Pythagorean theorem -THIS WILL HELP! (possibly)
    (Original post by PsychoAnalyser)
    Ok i'm out, sorry, I have to revise.
    Hey, I appreciate your replies
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    (Original post by Ilovemaths...)
    360-30 because its another revolution you add 180 degrees to 180 making 360 degrees. See in my working out, I do not get how that doesn't give you 1/2?
    Do you know CAST? How have you taught to use the quadrant diagram?

    For another revolution you need to add 360. So to get to that line in the first quadrant a second time you have to go all the way around anticlockwise (360) and then reach the line (+30) which gives you 360 + 30 = 390.
 
 
 
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