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Math Help - inequality

  1. #1
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    inequality

    Please help me with this question:



    Solve sin(-2x + \frac{\pi}{6}) = \frac{1}{2} for -\pi \leq x \leq 2\pi
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  2. #2
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    Hello,
    Quote Originally Posted by Musab View Post
    Please help me with this question:



    Solve sin(-2x + \frac{\pi}{6}) = \frac{1}{2} for -\pi \leq x \leq 2\pi
    -\pi \le x \le 2 \pi \implies -4 \pi \le -2x \le 2 \pi \implies -4 \pi +\frac \pi 6 \le -2x + \frac \pi 6 \le 2 \pi+\frac \pi 6
    Let X=-2x +\frac \pi 6

    So now, the problem is to solve \sin(X)=\frac 12 for -4 \pi +\frac \pi 6 \le X \le 2 \pi+\frac \pi 6

    If \sin(X)=a, then \sin \left(X+2k \pi\right)=a and \sin \left((2k+1) \pi -X\right)=a for any integer (positive or negative) k.

    Knowing that \sin \left(\frac \pi 6\right)=\frac 12, try to find all the solutions
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