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Math Help - Trigonometric Calculus

  1. #1
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    Trigonometric Calculus

    Okay, I always struggle with how to exactly work out the algebra for these trigonometric problems... The question is to identify where and what type of critical points lie on the function:

    f(x)=cos(2x)-1sin(x)

    between 0 and 2pi. Apparently there are 4.

    So far this is what I have done:
    f'(x)=-2sin(2x)-cosx

    0=-4sinxcosx-cosx

    cosx=-4sinxcosx

    -4sinx=1

    sinx=-1/4

    First of all I don't know how to solve for x in the final line without a calculator. Also, I only see that that would only possibly lead to two critical points between 0 and 2pi....

    Thank you anyone who can help me a bit!

    P.S. I kind of suck at this math formatting so forgive me...
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  2. #2
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    An equation on the form \cos{v} = c has the solutions v = v_0 + 2\pi n and v = -v_0 + 2\pi n

    An equation on the form \sin{v} = c has the solutions v = v_0 + 2\pi n and v = \pi-v_0 + 2\pi n

    In your case: \cos{x} = 0 or \sin{x} = -\frac{1}{4}
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  3. #3
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    When does sinx=-1/4... I can't find a special triangle or unit circle or graph to solve it...
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  4. #4
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    Quote Originally Posted by metaname90 View Post
    Okay, I always struggle with how to exactly work out the algebra for these trigonometric problems... The question is to identify where and what type of critical points lie on the function:

    f(x)=cos(2x)-1sin(x)
    Is that really what the problem said? That "1" in front of the sin(x) doesn't add anything and is peculiar.

    between 0 and 2pi. Apparently there are 4.

    So far this is what I have done:
    f'(x)=-2sin(2x)-cosx

    0=-4sinxcosx-cosx

    cosx=-4sinxcosx

    -4sinx=1
    If cos(x)\ne 0

    sinx=-1/4

    First of all I don't know how to solve for x in the final line without a calculator. Also, I only see that that would only possibly lead to two critical points between 0 and 2pi....

    Thank you anyone who can help me a bit!

    P.S. I kind of suck at this math formatting so forgive me...
    As Mondreus told you, either cos(x)= 0, which happens at x= \frac{\pi}{2} and at x= \frac{3\pi}{2} or sin(x)= -1/4 which happens at x= 3.3943 and at x= 6.0305, approximately. There is no "simple" form for that. Why do you want to solve it "without a calculator"?
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  5. #5
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    0 = -4\sin{x}\cos{x} - \cos{x}

    factoring ensures you do not miss possible solutions ...

    0 = -\cos{x}(4\sin{x} + 1)

    \cos{x} = 0 , \sin{x} = -\frac{1}{4}

    for 0 \le x < 2\pi ...

    x = \frac{\pi}{2} , x = \frac{3\pi}{2}

    x = 2\pi + \arcsin\left(-\frac{1}{4}\right)

    x = \pi - \arcsin\left(-\frac{1}{4}\right)
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  6. #6
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    I got the solutions now.... and yeah, the best I could do for sinx=-1/4 was 2pi+arcsin(-1/4) so I was wondering if I was missing out. In my calculus calculators are not permitted, in any of the quizzes or final exam, so I thought this was a strange solution.

    And whoever pointed out the "1" in front of sinx I know that is very peculiar, I don't know why I decided to copy it directly from the question, but that's definitely how it appeared.

    I found a video and then skeeter explained that you must factor in order to not lose answers, so that's all good.

    Thank you everyone!
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