inequality

**Musab** · September 5th 2008, 02:01 AM

Please help me with this question:

Solve $sin(-2x + \frac{\pi}{6}) = \frac{1}{2}$ for $-\pi \leq x \leq 2\pi$

**Moo** · September 5th 2008, 02:53 AM

Hello,

Originally Posted by Musab

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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