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Thread: Need help with a Trig equation

  1. #1
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    Need help with a Trig equation

    $\displaystyle 2sin\theta cos\theta-cos\theta=0$

    This is what I did

    First I took $\displaystyle cos\theta$ to other side resulting in $\displaystyle 2sin\theta cos\theta=cos\theta$
    Then I divided by $\displaystyle cos\theta$ resulting in $\displaystyle 2sin\theta =1$ then I divided by 2 $\displaystyle sin\theta = \frac{1}{2}$
    I am sure it's incorrect.. Can someone please help me
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  2. #2
    MHF Contributor
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    Hi

    Dividing by $\displaystyle \cos\theta$ makes you lose solutions.

    You must factor.

    $\displaystyle 2\sin\theta \cos\theta-\cos\theta=0$

    $\displaystyle \cos\theta (2\sin\theta-1)=0$

    leading to $\displaystyle \cos\theta = 0$ or $\displaystyle \sin\theta = \frac{1}{2}$
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  3. #3
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    Quote Originally Posted by daunder View Post
    $\displaystyle 2sin\theta cos\theta-cos\theta=0$

    This is what I did

    First I took $\displaystyle cos\theta$ to other side resulting in $\displaystyle 2sin\theta cos\theta=cos\theta$
    Then I divided by $\displaystyle cos\theta$ resulting in $\displaystyle 2sin\theta =1$ then I divided by 2 $\displaystyle sin\theta = \frac{1}{2}$
    I am sure it's incorrect.. Can someone please help me
    That would give you some solutions but not all of them. You can get them all by factorising:

    $\displaystyle cos(\theta)(2sin(\theta)-1) = 0$

    Then you can solve $\displaystyle cos(\theta) = 0$ to give $\displaystyle \theta = \frac{\pi}{2} \text { , } \frac{3\pi}{2}$

    Then solve $\displaystyle 2sin(\theta)-1 = 0$

    $\displaystyle \theta = \frac{\pi}{6} \text { , } \frac{5\pi}{6}$

    note: I have worked out the solutions of $\displaystyle 0 \leq \theta \leq 2\pi$
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  4. #4
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    Thanks for the quick reply
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  5. #5
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    Quote Originally Posted by daunder View Post
    $\displaystyle 2sin\theta cos\theta-cos\theta=0$

    This is what I did

    First I took $\displaystyle cos\theta$ to other side resulting in $\displaystyle 2sin\theta cos\theta=cos\theta$
    Then I divided by $\displaystyle cos\theta$ resulting in $\displaystyle 2sin\theta =1$ then I divided by 2 $\displaystyle sin\theta = \frac{1}{2}$
    I am sure it's incorrect.. Can someone please help me
    You can't divide by $\displaystyle \cos(\theta)$ as it can be equal to 0.

    Here is how to do.

    factorise by cos(theta) :

    $\displaystyle \cos \theta (2 \sin \theta-1)=0$

    the product ab=0 if and only if a=0 or b=0

    ----> $\displaystyle \cos \theta=0$ or $\displaystyle \sin \theta=\frac 12$


    For the first one... there are infinitely many solutions !
    we know that $\displaystyle \cos \frac \pi 2=0$
    but since cos(-x)=cos(x), we know that $\displaystyle \cos \left(-\frac \pi 2\right)=0$
    moreover, cos is a 2pi-periodic function.
    Hence $\displaystyle \boxed{\cos \left(\frac \pi 2 +2k\pi\right)=\cos \left(-\frac \pi 2 +2k\pi\right)=0}$
    these give all the possible solutions for cos(theta)=0

    (in general, if a is such that cos(a)=b, then the solutions to cos(x)=b are in the form a+2kpi or -a+2kpi, for any integer k)


    For the second one, there are infinitely many solutions too.
    we know that $\displaystyle \sin\left(\frac \pi 3\right)=0$
    but we know that $\displaystyle \sin(\pi-x)=\sin(x)$. Thus $\displaystyle \sin\left(-\frac \pi 3+\pi\right)=0$
    again, sin is a 2pi-periodic function.
    Hence $\displaystyle \sin\left(2k\pi\right)=\sin\left(-\frac \pi 3+\pi+2k\pi\right)=0$
    this can be rewritten :
    $\displaystyle \boxed{\sin\left(2k\pi\right)=\sin\left(-\frac \pi 3+(2k+1)\pi\right)=0}$


    Finally, the solutions are :
    $\displaystyle S=\left\{\tfrac{\pi}{2}+2k \pi \big| k \in \mathbb{Z}\right\} \cup \left\{-\tfrac{\pi}{2}+2k \pi \big| k \in \mathbb{Z}\right\} \cup \left\{2k \pi \big| k \in \mathbb{Z}\right\} \cup \left\{-\tfrac \pi 3+(2k+1)\pi \big| k \in \mathbb{Z}\right\}$

    does it look good to you ?




    Edit : woops, a bit late
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