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  1. #1
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    trigonometry question

    solve for x , -180<x<180

    2cos x+1= sin x

    This is what i did ..

    sin x -2 cos x=1

    \sqrt{5}\sin (x-63.43)=1

    so x-63.43=26.57

    x=90

    and x-63.43=-26.57 , -153.43

    x=36.86 , -90

    My answers are in red . But tat doesn't agree with the answer in the book . Thanks for helping .
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  2. #2
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    Quote Originally Posted by thereddevils View Post
    sin x -2 cos x=1

    \sqrt{5}\sin (x-63.43)=1

    so x-63.43=26.57
    How did you go from the first line to the second?
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  3. #3
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by thereddevils View Post
    solve for x , -180<x<180

    2cos x+1= sin x

    This is what i did ..

    sin x -2 cos x=1

    \sqrt{5}\sin (x-63.43)=1

    so x-63.43=26.57

    x=90

    and x-63.43=-26.57 , -153.43

    x=36.86 , -90

    My answers are in red . But tat doesn't agree with the answer in the book . Thanks for helping .
    2 cos x + 1 = sin x
    2 \cos x  + 1 = \sqrt{1 - \cos^2x}

    Squaring both sides, you get:

    4 \cos^2x+ 4 \cos x = 1 - \cos^2 x

    Move everything to one side:
    5\cos^2x + 4 \cos x - 1 = 0
    Sub: u = \cos x
    5u^2 + 4u -1 = 0

    u = \frac{-4 \pm 6}{10}

    giving -1 or 0.2.

    Set cosine x equal to each and solve. Check your solutions to make sure they work.

    Good luck!!
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  4. #4
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    Quote Originally Posted by Jameson View Post
    How did you go from the first line to the second?
    I used this r sin (x-a) , where r=\sqrt{1^2+2^2}=\sqrt{5}

    and tan a = 2


    do you see anything wrong in my working ?
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  5. #5
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    Quote Originally Posted by thereddevils View Post
    solve for x , -180<x<180

    2cos x+1= sin x

    This is what i did ..

    sin x -2 cos x=1

    \sqrt{5}\sin (x-63.43)=1

    so x-63.43=26.57

    x=90 Mr F says: This is correct.

    and x-63.43=-26.57 , -153.43 Mr F says: This is wrong. It should be x - 63.43 = 180 - 26.57 = 153.43. Note that sin(-A) = -sin(A) .....

    [snip]
    Therefore x - 63.43 = 153.43 => x = 216.86 degrees which is outside the domain therefore subtract 360 degrees to get the other value of x.
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