Please help me with this question:
Solve $\displaystyle sin(-2x + \frac{\pi}{6}) = \frac{1}{2}$ for $\displaystyle -\pi \leq x \leq 2\pi$
Hello,
$\displaystyle -\pi \le x \le 2 \pi \implies -4 \pi \le -2x \le 2 \pi \implies -4 \pi +\frac \pi 6 \le -2x + \frac \pi 6 \le 2 \pi+\frac \pi 6$
Let $\displaystyle X=-2x +\frac \pi 6$
So now, the problem is to solve $\displaystyle \sin(X)=\frac 12$ for $\displaystyle -4 \pi +\frac \pi 6 \le X \le 2 \pi+\frac \pi 6$
If $\displaystyle \sin(X)=a$, then $\displaystyle \sin \left(X+2k \pi\right)=a$ and $\displaystyle \sin \left((2k+1) \pi -X\right)=a$ for any integer (positive or negative) k.
Knowing that $\displaystyle \sin \left(\frac \pi 6\right)=\frac 12$, try to find all the solutions