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Math Help - more solving trig questions

  1. #1
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    more solving trig questions

    Hi
    Can someone help me solve the following equations, coz i have tired and i cannot understand how to do them.

    1)sin(2x)=cos(x)

    2)sin(8x)=cos(4x)

    P.S
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  2. #2
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    Quote Originally Posted by Paymemoney View Post
    Hi
    Can someone help me solve the following equations, coz i have tired and i cannot understand how to do them.

    1)sin(2x)=cos(x)

    2)sin(8x)=cos(4x)

    P.S
    1) \sin{2x} = \cos{x}

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

    2\sin{x}\cos{x} - \cos{x} = 0

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

    \cos{x} = 0 or 2\sin{x} - 1 = 0.


    Case 1: \cos{x} = 0

    x = \left\{\frac{\pi}{2}, \frac{3\pi}{2}\right\} + 2\pi n, where n is an integer representing the number of times you have gone around the unit circle...


    Case 2: 2\sin{x} - 1 = 0

    2\sin{x} = 1

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

    x = \left\{\frac{\pi}{6}, \frac{5\pi}{6}\right\} + 2\pi n.


    So x = \left\{\frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2}\right\} + 2\pi n.
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  3. #3
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    Quote Originally Posted by Paymemoney View Post
    Hi
    Can someone help me solve the following equations, coz i have tired and i cannot understand how to do them.

    1)sin(2x)=cos(x)

    2)sin(8x)=cos(4x)

    P.S
    2) \sin{8x} = \cos{4x}

    2\sin{4x}\cos{4x} = \cos{4x}

    Note that if we let 4x = X we have the exact same equation as in Q.1.

    So

    4x = \left\{\frac{\pi}{6},\frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2}\right\} + 2\pi n

    x = \left\{ \frac{\pi}{24}, \frac{\pi}{8}, \frac{5\pi}{24}, \frac{3\pi}{8} \right\} + \frac{\pi n}{2}
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