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Thread: Solving trig

  1. #1
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    Solving trig

    Need help on how solve the following equation for 0 <= x <= $\displaystyle 2\pi$:
    This is what i have done, but it is incorrect.

    $\displaystyle sin(2x)=cos(x)$
    $\displaystyle \frac{sin(2x)}{cos(x)}=0$
    $\displaystyle tan2x=0$which gets 0

    P.S
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  2. #2
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    Smile

    Quote Originally Posted by Paymemoney View Post
    Need help on how solve the following equation for 0 <= x <= $\displaystyle 2\pi$:
    This is what i have done, but it is incorrect.

    $\displaystyle sin(2x)=cos(x)$
    $\displaystyle \frac{sin(2x)}{cos(x)}=0$
    $\displaystyle tan2x=0$which gets 0

    P.S
    your second step is wrong and so what follows.
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  3. #3
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    Hello Paymemoney
    Quote Originally Posted by Paymemoney View Post
    Need help on how solve the following equation for 0 <= x <= $\displaystyle 2\pi$:
    This is what i have done, but it is incorrect.

    $\displaystyle sin(2x)=cos(x)$
    $\displaystyle \frac{sin(2x)}{cos(x)}=0$
    $\displaystyle tan2x=0$which gets 0

    P.S
    Noting that $\displaystyle \sin2x = 2\sin x \cos x$, we get:
    $\displaystyle 2\sin x \cos x = \cos x$
    You need to be a little careful now. Don't just divide both sides by $\displaystyle \cos x$, because it might be zero, and the first commandment of arithmetic is 'Thou shalt not divide by zero'. So you need to say:
    $\displaystyle 2\sin x \cos x = \cos x$

    $\displaystyle \Rightarrow \cos x = 0$ or $\displaystyle 2\sin x = 1$
    Can you go on to complete this now?
    Spoiler:
    (I get the answers $\displaystyle x = \frac{\pi}{6},x = \frac{\pi}{2},x = \frac{5\pi}{6},x = \frac{3\pi}{2}$.)


    Grandad
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  4. #4
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    thanks i get how to do it now
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