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Math Help - simpl inequation

  1. #1
    Super Member dhiab's Avatar
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    simpl inequation

    Solve in R :
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  2. #2
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    Smile

    hello
     <br />
2\sin (\frac{\pi}{6})=1<br />
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  3. #3
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    2 sin( 2x - pi/6 ) - 1 <= 0
    2 sin( 2x - pi/6 ) <= 1
    sin( 2x - pi/6 ) <= 1/2
    2x - pi/6 <= sin^-1(1/2)
    2x = sin^-1(1/2) + pi/6
    x = ( sin^-1(1/2) + pi/6 ) / 2

    Therefore x = 15.26...
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  4. #4
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    Quote Originally Posted by justmaths View Post
    2 sin( 2x - pi/6 ) - 1 <= 0
    2 sin( 2x - pi/6 ) <= 1
    sin( 2x - pi/6 ) <= 1/2
    2x - pi/6 <= sin^-1(1/2)
    2x = sin^-1(1/2) + pi/6
    x = ( sin^-1(1/2) + pi/6 ) / 2

    Therefore x = 15.26...
    There are several errors in this solution.
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  5. #5
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    Quote Originally Posted by Raoh View Post
    hello
     <br />
2\sin (\frac{\pi}{6})=1<br />
    @OP: Expanding a little on this hint:

    You can therefore write \sin \left(2x - \frac{\pi}{6}\right) = \sin \left(\frac{\pi}{6}\right).

    Now solve for x using the fact that if sin A = sin B then:

    Case 1: A = B + 2n \pi

    Case 2: A = (\pi - B) + 2n \pi

    where n is an integer.

    Now that you know the values of x for which 2 \sin \left(2x - \frac{\pi}{6}\right) - 1 = 0 you should be able to write down the intervals over which 2 \sin \left(2x - \frac{\pi}{6}\right) - 1 \leq 0.

    (drawing a graph might help ....)
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  6. #6
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    Quote Originally Posted by dhiab View Post
    Solve in R :
    An alternative approach:

    First solve 2 \sin \left(2x - \frac{\pi}{6}\right) - 1 = 0:

    \sin \left(2x - \frac{\pi}{6}\right) = \frac{1}{2}

    Case 1: 2x - \frac{\pi}{6} = \frac{\pi}{6} + 2n \pi

    Case 2: 2x - \frac{\pi}{6} = \frac{5\pi}{6} + 2n \pi

    where n is an integer.

    Now solve for x and then solve the inequality as explained in my previous post.
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  7. #7
    Super Member dhiab's Avatar
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    Quote Originally Posted by mr fantastic View Post
    An alternative approach:

    First solve 2 \sin \left(2x - \frac{\pi}{6}\right) - 1 = 0:

    \sin \left(2x - \frac{\pi}{6}\right) = \frac{1}{2}

    Case 1: 2x - \frac{\pi}{6} = \frac{\pi}{6} + 2n \pi

    Case 2: 2x - \frac{\pi}{6} = \frac{5\pi}{6} + 2n \pi

    where n is an integer.

    Now solve for x and then solve the inequality as explained in my previous post.
    Hello : You can solve the inequality by trigo-circle Thank you
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  8. #8
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    Quote Originally Posted by dhiab View Post
    Hello: you have inequation et not equation Thank you
    Yes, we understood that! Apparently you did not understand the solution. Since the functions here are continuous, they can only change from "<" to ">" and vice-versa at "=". First solve the equation, then, as mr. fantastic said, check the intervals between the solutions to the equation to see where it is ">" and "<".
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