# How do I solve sin(5x-pi/3)<=0, [-pi/2,pi/2]?

Apr 8, 2018

Given: $\sin \left(5 x - \frac{\pi}{3}\right) \le 0 , \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$

First, do this without the domain restriction.

$\sin \left(5 x - \frac{\pi}{3}\right) \le 0$

This is true when the primary values of the argument are:

$- \pi \le 5 x - \frac{\pi}{3} \le 0$

Add $\pi 3$ to all 3:

$- \frac{2 \pi}{3} \le 5 x \le \frac{\pi}{3}$

Add integer multiples of $2 \pi$

$- \frac{2 \pi}{3} + 2 \pi n \le 5 x \le \frac{\pi}{3} + 2 \pi n , n \in \mathbb{Z}$

Divide everything by 5:

$- \frac{2 \pi}{15} + \frac{2}{5} \pi n \le x \le \frac{\pi}{15} + \frac{2}{5} \pi n , n \in \mathbb{Z}$

I have graphed the domain restriction: $\textcolor{b l u e}{- \frac{\pi}{2} \le x \le \frac{\pi}{2}}$ simultaneously with $\textcolor{red}{- \frac{2 \pi}{15} + \frac{2}{5} \pi n \le x \le \frac{\pi}{15} + \frac{2}{5} \pi n , n \in \mathbb{Z}}$:

Please observe that there are 3 values of n, -1, 0 and 1, that fall within the domain but the leftmost (n = -1) will be truncated by the domain's lower bound:

n = -1:

$- \frac{\pi}{2} \le x \le \frac{\pi}{15} - \frac{2}{5} \pi$

$n = 0$

$- \frac{2 \pi}{15} \le x \le \frac{\pi}{15}$

$n = 1$

$- \frac{2 \pi}{15} + \frac{2}{5} \pi \le x \le \frac{\pi}{15} + \frac{2}{5} \pi$

The union of the 3 domain segments is the solution to the inequality with the domain restriction.