Answer
$\left\{\frac{\pi}{12}, \frac{7\pi}{12}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{17\pi}{12} \frac{23\pi}{12}\right\}$
Work Step by Step
Add $1$ to both sides:
$\sqrt2 \cdot \cos{(3x)}=1$
Divide $\sqrt2$ to both sides:
$\cos{(3x)}=\frac{1}{\sqrt2}
\\\cos{(3x)}=\frac{\sqrt2}{2}
\\3x=\cos^{-1}{(\frac{\sqrt2}{2})}$
Use a scientific calculator's inverse cosine function to obtain:
$3x=\frac{\pi}{4} \text{ or } \frac{7\pi}{4} \text{ or } \frac{9\pi}{4} \text{ or } \frac{15\pi}{4} \text{ or } \frac{17\pi}{4} \text{ or } \frac{23\pi}{4}
\\x=\frac{\pi}{12} \text{ or } \frac{7\pi}{12} \text{ or } \frac{9\pi}{12} \text{ or } \frac{15\pi}{12} \text{ or } \frac{17\pi}{12} \text{ or } \frac{23\pi}{12}
\\x=\frac{\pi}{12} \text{ or } \frac{7\pi}{12} \text{ or } \frac{3\pi}{4} \text{ or } \frac{5\pi}{4} \text{ or } \frac{17\pi}{12} \text{ or } \frac{23\pi}{12}$
