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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 -\pi \leq x \leq \pi,

    \cos^2x = sinxcosx

    Using my graphing calculator I get four solutions.

    Dividing through by cos^2x I get:

    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 -\pi \leq x \leq \pi,
    \cos^2(x) = \sin(x)\cos(x)
    That can be written as \cos(x)(\cos(x)-\sin(x))=0.

    Now solve these two \cos(x)=0~\&~\cos(x)=\sin(x)~.
    Thanks from Furyan
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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 -\pi \leq x \leq \pi,

    \cos^2x = sinxcosx

    Using my graphing calculator I get four solutions.

    Dividing through by cos^2x I get:

    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
    cos^2(x) = sin(x)!cos(x)

    Factoring
    [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 -\pi \leq x \leq \pi,

    \cos^2x = sinxcosx

    Using my graphing calculator I get four solutions.

    Dividing through by cos^2x I get:

    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 \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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