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Thread: cos^2x=sinxcosx

  1. #1
    Member Furyan's Avatar
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    cos^2x=sinxcosx

    Hello

    I am struggling with a simple trig equation, again.

    The question is to solve, on the interval $\displaystyle -\pi \leq x \leq \pi$,

    $\displaystyle \cos^2x = sinxcosx$

    Using my graphing calculator I get four solutions.

    Dividing through by $\displaystyle cos^2x$ I get:

    $\displaystyle tanx = 1$, for which there are only two solutions. Two of the four I should have.

    I have come across this before, where decreasing the power reduces the number of solutions, which makes sense, although I don't understand it. I'd by grateful if someone would explain where I'm going wrong.

    Thank you
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  2. #2
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    Re: cos^2x=sinxcosx

    Quote Originally Posted by Furyan View Post
    The question is to solve, on the interval $\displaystyle -\pi \leq x \leq \pi$,
    $\displaystyle \cos^2(x) = \sin(x)\cos(x)$
    That can be written as $\displaystyle \cos(x)(\cos(x)-\sin(x))=0$.

    Now solve these two $\displaystyle \cos(x)=0~\&~\cos(x)=\sin(x)~.$
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  3. #3
    Forum Admin topsquark's Avatar
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    Re: cos^2x=sinxcosx

    Quote Originally Posted by Furyan View Post
    Hello

    I am struggling with a simple trig equation, again.

    The question is to solve, on the interval $\displaystyle -\pi \leq x \leq \pi$,

    $\displaystyle \cos^2x = sinxcosx$

    Using my graphing calculator I get four solutions.

    Dividing through by $\displaystyle cos^2x$ I get:

    $\displaystyle tanx = 1$, for which there are only two solutions. Two of the four I should have.

    I have come across this before, where decreasing the power reduces the number of solutions, which makes sense, although I don't understand it. I'd by grateful if someone would explain where I'm going wrong.

    Thank you
    $\displaystyle cos^2(x) = sin(x)!cos(x)$

    Factoring
    $\displaystyle [cos(x)] \cdot cos(x) = [cos(x)] \cdot sin(x)$

    Notice that there will be solutions when
    cos(x) = 0
    and when
    cos(x) = sin(x)

    -Dan
    Thanks from Furyan
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  4. #4
    Member Furyan's Avatar
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    Re: cos^2x=sinxcosx

    Hello Plato and topsquark,

    Thank you both very much. I actually understand that now and will look out for it in the future.
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  5. #5
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    Re: cos^2x=sinxcosx

    Quote Originally Posted by Furyan View Post
    Hello

    I am struggling with a simple trig equation, again.

    The question is to solve, on the interval $\displaystyle -\pi \leq x \leq \pi$,

    $\displaystyle \cos^2x = sinxcosx$

    Using my graphing calculator I get four solutions.

    Dividing through by $\displaystyle cos^2x$ I get:

    $\displaystyle tanx = 1$, for which there are only two solutions. Two of the four I should have.

    I have come across this before, where decreasing the power reduces the number of solutions, which makes sense, although I don't understand it. I'd by grateful if someone would explain where I'm going wrong.

    Thank you
    The reason you ended up with less solutions than you should have is because $\displaystyle \displaystyle \begin{align*} \cos^2{x} \end{align*}$ CAN equal 0. You can not divide by 0.
    Thanks from MarkFL and Furyan
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