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Math Help - Solving for unknown "x" involving trig function

  1. #1
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    Post Solving for unknown "x" involving trig function

    I have been trying to solve 3 questions, I simply cant. Please find the attachment.

    Question
    13e: I am not sure whether I did it right or not
    13g: After putting it in quadratic formula, i dont know how to deal with the square-root part
    13h: I just cant seem to simplify the left side and solve for X

    Thanks, all help is appreciated

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  2. #2
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    Quote Originally Posted by dennosan View Post
    13g: After putting it in quadratic formula, i dont know how to deal with the square-root part

    \displaystyle\sin x = \frac{-(-3)\pm\sqrt{(-3)^2- 4(1)(-2)}}{2(1)}

    \displaystyle\sin x = \frac{3\pm\sqrt{9+8}}{2}

    \displaystyle\sin x = \frac{3\pm\sqrt{17}}{2}

    \displaystyle\sin x = \frac{3+\sqrt{17}}{2}, \frac{3-\sqrt{17}}{2}

    discarding the first value

    \displaystyle x = \sin^{-1}\left( \frac{3-\sqrt{17}}{2}\right) = \dots
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  3. #3
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    For problems like 13e, I always write everything in terms of sin's and cos's:

    \cot(x)\cos^{2}(x)=2\cot(x) implies

    \displaystyle{\frac{\cos(x)}{\sin(x)}\,\cos^{2}(x)  =2\frac{\cos(x)}{\sin(x)}.}

    So we see immediately that locations where \sin(x)=0 are not allowed, which means x\not=n\pi, for all n\in\mathbb{Z}.

    Also, we wish to record the fact that places where \cos(x)=0 are a solution. So that occurs at odd integer multiples of \pi/2, so at x=(2n+1)\pi/2. Now that we have recorded those two facts, we may cancel all kinds of stuff. What do you get when you do that?

    For 13h, I would notice that secant squared's look like tangent squared's through the Pythagorean Theorem:

    \sin^{2}(x)+\cos^{2}(x)=1 implies, through dividing through by \cos^{2}(x), that

    \tan^{2}(x)+1=\sec^{2}(x).

    So what can you do then?
    Last edited by Ackbeet; August 27th 2010 at 05:06 AM. Reason: Square the tangent.
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  4. #4
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    d is wrong
    wrong factorisation.
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