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Math Help - solving for angle in equation

  1. #1
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    solving for angle in equation

    am trying to find the value for x between 0 and 360 degress. but i think am doing something wrong.

    2-sinx = cos^2x + 7sin^2x

    2-sinx = 1-sin^2x + 7sin^2x

    2-sinx = 1- 6sin^2x

    can someone help me to solve this am stuck here..
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  2. #2
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    Quote Originally Posted by sigma1 View Post
    am trying to find the value for x between 0 and 360 degress. but i think am doing something wrong.

    2-sinx = cos^2x + 7sin^2x

    2-sinx = 1-sin^2x + 7sin^2x

    \color{red}{2-sinx = 1- 6sin^2x}

    can someone help me to solve this am stuck here..
    Dear sigma1,

    There is a mistake in the highlighted line. Can you find it?
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  3. #3
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    Quote Originally Posted by Sudharaka View Post
    Dear sigma1,

    There is a mistake in the highlighted line. Can you find it?

    yes i see the error.

    it should be

    2-sinx = 1+6sin^2x
    1-sinx = 6sin^2x
    -sinx -6sin^2x=-1
    -sinx(1+6sinx)=-1

    where do i go from here?
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  4. #4
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    Quote Originally Posted by sigma1 View Post
    am trying to find the value for x between 0 and 360 degress. but i think am doing something wrong.

    2-sinx = cos^2x + 7sin^2x

    2-sinx = 1-sin^2x + 7sin^2x

    2-sinx = 1 + 6sin^2x

    can someone help me to solve this am stuck here..
    So you will have:

    6sin^2x + sinx -1 =0

    let  u = sinx and you can see that this is a quadratic equation:

    6u^2 +u -1 = 0

    Now solve for u using the quadratic equation formula, which will give you the value of u, that is sinx.

    after getiing sinx, you can calculate the angle(x)
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  5. #5
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by sigma1 View Post
    yes i see the error.

    it should be

    2-sinx = 1+6sin^2x
    1-sinx = 6sin^2x
    -sinx -6sin^2x=-1
    -sinx(1+6sinx)=-1

    where do i go from here?
    this is not going to work. refer to the above post and work on the quadratic equation
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  6. #6
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    Quote Originally Posted by harish21 View Post
    So you will have:

    6sin^2x + sinx -1 =0

    let  u = sinx and you can see that this is a quadratic equation:

    6u^2 +u -1 = 0

    Now solve for u using the quadratic equation formula, which will give you the value of u, that is sinx.

    after getiing sinx, you can calculate the angle(x)
    This equation will also factorise

    6\sin ^2(x) + \sin(x) -1 = (3\sin(x)-1)(2\sin(x)+1)=0
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  7. #7
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    Quote Originally Posted by harish21 View Post
    this is not going to work. refer to the above post and work on the quadratic equation
    ok thanks for that. i solved the equation and got 2 values for u.

    u =1/3 and u = - 1/2

    which one of the values should i use.

    i used -1/2 though to get 6sin^ -[\frac{1}{2}] - 1 = 0

    sin^2x = \frac{1}{2}
    sinx = 1/2
    x = sin^{-1}  1/2
    x = 30
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  8. #8
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    Hi sigma1,

    the quadratic equation allowed you to discover the 2 possible values for sinx

    Now you need to find the valid values of the angle x.

    sinx may be conveniently thought of as the vertical co-ordinate
    of a point on a circle centred at (0,0) with radius 1.

    Then you can see, using a sketch that a vertical co-ordinate of \frac{1}{3}

    gives 2 possible angles, x=sin^{-1}\left(\frac{1}{3}\right)

    and 180^o-sin^{-1}\left(\frac{1}{3}\right)


    For sinx=-\frac{1}{2} on the vertical axis,

    you can calculate the acute angle sin^{-1}\left(\frac{1}{2}\right)

    to get the acute angle the point on the circle makes under the x-axis.

    Then subtract this from 360 degrees and add it to 180 degrees.

    You really need to understand that procedure
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  9. #9
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    Quote Originally Posted by Archie Meade View Post
    Hi sigma1,

    the quadratic equation allowed you to discover the 2 possible values for sinx

    Now you need to find the valid values of the angle x.

    sinx may be conveniently thought of as the vertical co-ordinate
    of a point on a circle centred at (0,0) with radius 1.

    Then you can see, using a sketch that a vertical co-ordinate of \frac{1}{3}

    gives 2 possible angles, x=sin^{-1}\left(\frac{1}{3}\right)

    and 180^o-sin^{-1}\left(\frac{1}{3}\right)


    For sinx=-\frac{1}{2} on the vertical axis,

    you can calculate the acute angle sin^{-1}\left(\frac{1}{2}\right)

    to get the acute angle the point on the circle makes under the x-axis.

    Then subtract this from 360 degrees and add it to 180 degrees.

    You really need to understand that procedure

    i am a bit unclear on the theory you trying to bring across to me... its the first am seeing something like this... could i have missed this in class?
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  10. #10
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    Quote Originally Posted by sigma1 View Post
    i am a bit unclear on the theory you trying to bring across to me... its the first am seeing something like this... could i have missed this in class?
    quite possibly,
    it may not always be brought across as

    sin(angle)=y\ co-ordinate

    cos(angle)=x\ co-ordinate

    in a circle, centre (0,0) and radius 1

    However, i feel it's quite easy to work with that way.

    You may instead be presented with

    sin is + in the 1st and 2nd quadrants, sin is - in the 3rd and 4th.
    cos is + in the 1st and 4th quadrants, cos is - in the 2nd and 3rd.

    Hence for your positive answer for sinx,
    this corresponds to one angle in the 1st quadrant, 0 to 90 degrees
    and another angle in the 2nd quadrant, 90 to 180 degrees.

    Then work from there knowing that the angle in the 2nd quadrant is 180-(angle in 1st)

    Similarly for cos, using the guidelines for that.

    i much prefer the horizontal and vertical co-ordinates way,
    after all that's why we can give sin and cos their polarities in the 4 quadrants.

    Whichever way you do it, remember sinx points out 2 angles as does cosx.
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  11. #11
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    Quote Originally Posted by sigma1 View Post
    ok thanks for that. i solved the equation and got 2 values for u.

    u =1/3 and u = - 1/2

    which one of the values should i use.

    i used -1/2 though to get 6sin^ -[\frac{1}{2}] - 1 = 0

    sin^2x = \frac{1}{2}
    sinx = 1/2
    x = sin^{-1}  1/2
    x = 30
    So your values are u = \frac{1}{3} and u =\frac{-1}{2}, which you got correct.

    That means:

    sinx = \frac{1}{3}
    or
    sinx=\frac{-1}{2}

    now calculate x.

    Note that your question is asking for x values between 0 to 360 degrees.
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  12. #12
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    so i can say that this answer is 19. 47 degrees
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  13. #13
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by sigma1 View Post
    so i can say that this answer is 19. 47 degrees
    yes
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