Originally Posted by
yeongil Are there any restrictions on x? Usually we solve with the restriction $\displaystyle 0 \le x < 2\pi$. If that's the case, then we'll need to expand on pickslides' work a bit. If we have the restriction $\displaystyle 0 \le x < 2\pi$, then it follows that $\displaystyle 0 \le {\color{red}3}x < {\color{red}6}\pi$.
Between 0 and 2 pi we also have 5pi/3 as a possible answer:
$\displaystyle 3x = \frac{5\pi}{3}$
For each answer, we need to find coterminal angles to it that is in $\displaystyle 0 \le 3x < 6\pi$, so we add 2pi and 4pi to each:
$\displaystyle 3x = \frac{\pi}{3}$
$\displaystyle 3x = \frac{\pi}{3} + 2\pi = \frac{7\pi}{3}$
$\displaystyle 3x = \frac{\pi}{3} + 4\pi = \frac{13\pi}{3}$
$\displaystyle 3x = \frac{5\pi}{3}$
$\displaystyle 3x = \frac{5\pi}{3} + 2\pi = \frac{11\pi}{3}$
$\displaystyle 3x = \frac{5\pi}{3} + 4\pi = \frac{17\pi}{3}$
Now take each equation and divide both sides by 3 to solve for x. So we'll get
$\displaystyle x = \frac{\pi}{9},\;\frac{5\pi}{9},\;\frac{7\pi}{9},\; \frac{11\pi}{9},\;\frac{13\pi}{9},\;\frac{17\pi}{9 },\;$
These are all possible values of x that is in $\displaystyle 0 \le x < 2\pi$.
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