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Math Help - Solve sin(x) = cos(2x) for all x

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    Solve sin(x) = cos(2x) for all x

    Hello,

    The question is: Solve sin(x) = cos(2x) for all x.

    My solution:
    sin(x) = cos(2x)
    sin(x) = sin(1/2 pi - 2x)
    x = 1/2 pi - 2x
    3x = 1/2 pi
    x = 1/6 pi

    However, the question says "solve for all x". I understand that fact that there are multiple x-es since both functions are periodic. But how do I find the other x-es? Please help!
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    Re: Solve sin(x) = cos(2x) for all x

    Quote Originally Posted by Lotte1990 View Post
    Hello,

    The question is: Solve sin(x) = cos(2x) for all x.

    My solution:
    sin(x) = cos(2x)
    sin(x) = sin(1/2 pi - 2x)
    x = 1/2 pi - 2x
    3x = 1/2 pi
    x = 1/6 pi

    However, the question says "solve for all x". I understand that fact that there are multiple x-es since both functions are periodic. But how do I find the other x-es? Please help!
    It would be better to use the identity \displaystyle \begin{align*} \cos{2x} \equiv 1 - 2\sin^2{x} \end{align*}, so that your equation becomes

    \displaystyle \begin{align*}  \sin{x} &= \cos{2x} \\ \sin{x} &= 1 - 2\sin^2{x} \\ 2\sin^2{x} + \sin{x} - 1 &= 0 \end{align*}

    Now solve this quadratic.
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    Re: Solve sin(x) = cos(2x) for all x

    2\sin^2{x} + \sin{x} - 1 = 0

    \sin{x} = p

    2p^2 + p - 1 = 0

    p^2 + \frac{1}{2}p - \frac{1}{2} = 0

    p = 1 or p = -\frac{1}{2}

    \sin{x} = 1 or \sin{x} = -\frac{1}{2}

    x = \arcsin{1} or x = \arcsin{-\frac{1}{2}}

    x = \frac{1}{2}\pi + 2k\pi or x = -\frac{1}{6}\pi + 2k\pi or x = -\frac{5}{6}\pi + 2k\pi in which k is an integer.

    Is this correct? And is this the COMPLETE answer? Thanks!
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    Re: Solve sin(x) = cos(2x) for all x

    Yes, that is the complete answer.
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    Re: Solve sin(x) = cos(2x) for all x

    Ok, then please explain to me why I find x equal to 1/6 pi in the first post, but not in the second one. I don't understand...
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    Re: Solve sin(x) = cos(2x) for all x

    solutions for the interval [0, 2\pi] ...

    \sin{x} = \cos(2x)

    \sin{x} = 1 - 2\sin^2{x}

    2\sin^2{x} + \sin{x} - 1 = 0

    (2\sin{x} - 1)(\sin{x} + 1) = 0

    \sin{x} = \frac{1}{2} ... x = \frac{\pi}{6} , \frac{5\pi}{6}

    \sin{x} = -1 ... x = \frac{3\pi}{2}
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    Re: Solve sin(x) = cos(2x) for all x

    Quote Originally Posted by Lotte1990 View Post
    Ok, then please explain to me why I find x equal to 1/6 pi in the first post, but not in the second one. I don't understand...
    because you assumed that if sin(x) = sin(pi/2 - 2x), that x = pi/2 - 2x. but this is not true, sin(x) is not a 1-1 function.

    for example, sin(pi/2 - 2x) = sin(5pi/2 - 2x) as well (leading to x = 5pi/6), and also cos(2x) = cos(-2x), so

    we have sin(x) = sin(pi/2 + 2x) also (leading to x = -pi/2, and thus (by adding 2pi) x = 3pi/2).

    it is customary to find the solutions "in one revolution" (either from -pi to pi, or 0 to 2pi), and then add 2kpi to the results.
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    Re: Solve sin(x) = cos(2x) for all x

    Quote Originally Posted by Lotte1990 View Post
    2\sin^2{x} + \sin{x} - 1 = 0

    \sin{x} = p

    2p^2 + p - 1 = 0

    p^2 + \frac{1}{2}p - \frac{1}{2} = 0

    p = 1 or p = -\frac{1}{2}

    \sin{x} = 1 or \sin{x} = -\frac{1}{2}

    x = \arcsin{1} or x = \arcsin{-\frac{1}{2}}

    x = \frac{1}{2}\pi + 2k\pi or x = -\frac{1}{6}\pi + 2k\pi or x = -\frac{5}{6}\pi + 2k\pi in which k is an integer.

    Is this correct? And is this the COMPLETE answer? Thanks!
    You have solved the quadratic equation incorrectly.

    \displaystyle \begin{align*} 2p^2 + p - 1 &= 0 \\ 2p^2 + 2p - p - 1 &= 0 \\ 2p(p + 1) -1(p + 1) &= 0 \\ (p + 1)(2p - 1) &= 0 \\ p + 1 = 0 \textrm{ or } 2p - 1 &= 0 \\ p = -1 \textrm{ or } p &= \frac{1}{2} \\ \sin{x} = -1 \textrm{ or }\sin{x} &= \frac{1}{2} \end{align*}
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