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Math Help - [SOLVED] Trigonometry

  1. #1
    Member looi76's Avatar
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    [SOLVED] Trigonometry

    Question:
    Find all the values of x in the interval 0^o \leq x \leq 180^o which satisfy the equation \sin{3x} + 2\cos{3x} = 0.

    Attempt:

    \sin{3x} + 2\cos{3x} = 0.

    \sin{3x} = -2\cos{3x}

    \frac{\sin{3x}}{\cos{3x}} = -2

    \tan{3x} = -2

    3x = \tan^{-1}(-2)

    Calculator gives no value for \tan^{-1}(-2), need help.
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  2. #2
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    Use \; \tan \left( -x \right) = - \tan \left( x \right). Same is true for the inverse.
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    It gives a value for arctan(-2), but it gives a negative result..

    Also, I intended to substitute \sin(3x) by \sqrt{1-\cos^2(3x)}, but I don't know what it'll give
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  4. #4
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    Quote Originally Posted by looi76 View Post
    Question:
    Find all the values of x in the interval 0^o \leq x \leq 180^o which satisfy the equation \sin{3x} + 2\cos{3x} = 0.

    Attempt:

    \sin{3x} + 2\cos{3x} = 0.

    \sin{3x} = -2\cos{3x}

    \frac{\sin{3x}}{\cos{3x}} = -2

    \tan{3x} = -2

    3x = \tan^{-1}(-2)

    Calculator gives no value for \tan^{-1}(-2), need help.
    Quote Originally Posted by xifentoozlerix View Post
    Use \; \tan \left( -x \right) = - \tan \left( x \right). Same is true for the inverse.
    \tan{3x} = -2

    -\tan{3x} = 2

    -3x = \tan^{-1}(2)

    x = \frac{\tan^{-1}(2)}{-3}

    x = 158^o

    Where did I go wrong?
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    x = \frac{{\tan ^{ - 1} \left( 2 \right)}}{{ - 3}}\approx\frac{{1.10715}}{{ - 3}} \approx  - 0.36905

    Now you can see by the original equation that the function has frequency 3, which means that it's value repeats every \frac{2\pi}{3} radians or every 120^\circ.
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  6. #6
    Member looi76's Avatar
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    Quote Originally Posted by xifentoozlerix View Post
    x = \frac{{\tan ^{ - 1} \left( 2 \right)}}{{ - 3}}\approx\frac{{1.10715}}{{ - 3}} \approx  - 0.36905

    Now you can see by the original equation that the function has frequency 3, which means that it's value repeats every \frac{2\pi}{3} radians or every 120^\circ.
    My mistake was that I didn't configure the calculator to radian mode. It was kept on degree mode. When should I choose the degree mode and when should I choose the radian mode?
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  7. #7
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    Quote Originally Posted by looi76 View Post
    Question:
    Find all the values of x in the interval 0^o \leq x \leq 180^o which satisfy the equation \sin{3x} + 2\cos{3x} = 0.

    Attempt:

    \sin{3x} + 2\cos{3x} = 0.

    \sin{3x} = -2\cos{3x}

    \frac{\sin{3x}}{\cos{3x}} = -2

    \tan{3x} = -2

    3x = \tan^{-1}(-2)

    Calculator gives no value for \tan^{-1}(-2), need help.
    Quote Originally Posted by xifentoozlerix View Post
    x = \frac{{\tan ^{ - 1} \left( 2 \right)}}{{ - 3}}\approx\frac{{1.10715}}{{ - 3}} \approx  - 0.36905

    Now you can see by the original equation that the function has frequency 3, which means that it's value repeats every \frac{2\pi}{3} radians or every 120^\circ.
    and can you please tell me all the possible values of x in the interval 0^o \leq x \leq 180^o
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  8. #8
    Moo
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    Add 120 to -0.369, it'll be between 0 and 180
    If you add 120 once again, it'll go over 180. So the only solution is 120-0.369
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  9. #9
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    I just realized that the roots repeat in \frac{2\pi}{6}=\frac{\pi}{3}. So you will have answers of x\approx \left[ -0.36905+\frac{\pi}{3},-0.36905+\frac{2\pi}{3},-0.36905+\pi \right]
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  10. #10
    Moo
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    Thought the period was 2pi, but it's pi...

    So add each time pi/3, that is to say 60
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  11. #11
    Member looi76's Avatar
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    But the answer given for the values of x in the textbook is:
    x = 38.9 , x = 98.9 , x = 158.9

    Can you please explain?
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  12. #12
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    Quote Originally Posted by looi76 View Post
    But the answer given for the values of x in the textbook is:
    x = 38.9 , x = 98.9 , x = 158.9

    Can you please explain?
    Yes that is correct. That is because the others have answered in "radians" and your book has answers in degrees.
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  13. #13
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    Can I solve this question without using Radians at all?
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    Ok, let's do it again

    (in degrees)

    3x=\tan^{-1} (-2) \approx -63.4

    x \approx -21.1

    Then, add 60 as far as you remain between 0 and 180

    It should give you the values you want
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  15. #15
    Lord of certain Rings
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    Quote Originally Posted by looi76 View Post
    Can I solve this question without using Radians at all?
    Yes. You had previously got the answer as 158.9 degrees.

    Now add or subtract 60 degrees to get all other answers.
    Adding 60 degrees takes it above 180 degrees, so do not do it.

    However

    158.9 - 60 = 98.9

    98.9 - 60 = 38.9
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