# inequality

• Sep 5th 2008, 02:01 AM
Musab
inequality

Solve $sin(-2x + \frac{\pi}{6}) = \frac{1}{2}$ for $-\pi \leq x \leq 2\pi$
• Sep 5th 2008, 02:53 AM
Moo
Hello,
Quote:

Originally Posted by Musab
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 (Wink)