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Math Help - trigonometric problem

  1. #1
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    trigonometric problem

    please help me with this:

    solve the following equation for 0<x<360.

    a) 4cot^2 x =5 cos x




    b) 4cos^2 x =9-2 sec^2 x
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  2. #2
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    Hey, when attempting these questions, do not cancel out anything

    Here's a start for the first question:

     4cot^2x = 5cos x

     4\frac{\cos^2x}{\sin^2x} - 5\cos x = 0

     \cos x \left(\frac{4\cos x - 5(1-\cos^2x)}{\sin^2x} \right) = 0

    Solve each term for 0.
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  3. #3
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    Quote Originally Posted by Gusbob View Post
    Hey, when attempting these questions, do not cancel out anything

    Here's a start for the first question:

     4cot^2x = 5cos x

    still dont get....can give me step by step?

     4\frac{\cos^2x}{\sin^2x} - 5\cos x = 0

     \cos x \left(\frac{4\cos x - 5(1-\cos^2x)}{\sin^2x} \right) = 0

    Solve each term for 0.

    still dont know..can u teach me step by step sir?
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  4. #4
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    Hello mastermin346
    Quote Originally Posted by mastermin346 View Post
    please help me with this:

    solve the following equation for 0<x<360.

    a) 4cot^2 x =5 cos x
    4\cot^2 x =5 \cos x

    \Rightarrow \dfrac{4\cos^2x}{\sin^2x}=5\cos x
    You'll notice that you now have a factor of \cos x on both sides of the equation. So one solution is
    \cos x = 0
    (which you can solve for 0<\<360^o)

    But if \cos x \ne 0 we can divide both sides by \cos x to get:
    \dfrac{4\cos x}{\sin^2x}=5

    \Rightarrow 4\cos x = 5\sin^2x
    =5-5\cos^2x
    Now re-arrange this as a quadratic equation in \cos x, and solve.

    b) 4cos^2 x =9-2 sec^2 x
    If we put \cos^2x = c, the equation can be re-written:
    4c=9-\dfrac{2}{c}
    Multiply both sides by c, and re-arrange this as a quadratic. Solve for c and hence \cos x.

    Can you complete them both now?

    Grandad
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