solving trig equation

First off, is there any restrictions for x? Like $0 \le x < 2\pi$ or something? Because that's what I'm going to assume.

You're thinking too hard. Ask your self, the tangent of what angle is 1? It's one of the special angles, and you know that it's $\frac{\pi}{4}$ . Tangent is positive in the 3rd quadrant, so another angle would be $\frac{\pi}{4}$ . But we're not done.

What we found is that ${\color{red}4}x = \frac{\pi}{4}$ and ${\color{red}4}x = \frac{5\pi}{4}$ . But if we go by the restriction $0 \le x < 2\pi$ , then it follows that $0 \le {\color{red}4}x < {\color{red}8}\pi$ . This means that there are a lot more angles whose tangent is 1. Since the period of tangent is $\pi$ , we can just take $4x = \frac{\pi}{4}$ and keep adding $\pi$ until we find all angles in $0 \le {\color{red}4}x < {\color{red}8}\pi$ . The complete list of angles is therefore
$4x = \frac{\pi}{4},\;\frac{5\pi}{4},\;\frac{9\pi}{4},\; \frac{13\pi}{4},\;\frac{17\pi}{4},\;\frac{21\pi}{4 },\;\frac{25\pi}{4},\;\frac{29\pi}{4}$ .

Now take each angle and divide by 4 to get our x values:
$x = \frac{\pi}{16},\;\frac{5\pi}{16},\;\frac{9\pi}{16} ,\;\frac{13\pi}{16},\;\frac{17\pi}{16},\;\frac{21\ pi}{16},\;\frac{25\pi}{16},\;\frac{29\pi}{16}$ .

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