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Thread: trigonometric problem

  1. #1
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    trigonometric problem

    please help me with this:

    solve the following equation for $\displaystyle 0<x<360$.

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




    b)$\displaystyle 4cos^2 x =9-2 sec^2 x$
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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:

    $\displaystyle 4cot^2x = 5cos x $

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

    $\displaystyle \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:

    $\displaystyle 4cot^2x = 5cos x $

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

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

    $\displaystyle \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 $\displaystyle 0<x<360$.

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

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

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

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

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

    Can you complete them both now?

    Grandad
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