Answer:
[tex]\large\boxed{\Phi=\dfrac{\pi}{9}+\dfrac{2k\pi}{3}\ or\ \Phi=-\dfrac{\pi}{9}+\dfrac{2k\pi}{3}}[/tex]
Step-by-step explanation:
[tex]\cos3\Phi=\dfrac{1}{2}\qquad\text{substitute}\ 3\Phi=\theta\\\\\cos\theta=\dfrac{1}{2}\iff\theta=\dfrac{\pi}{3}+2k\pi\ or\ \theta=-\dfrac{\pi}{3}+2k\pi\qquad k\in\mathbb{Z}\\\\\text{We're going back to substitution:}\\\\3\Phi=\dfrac{\pi}{3}+2k\pi\ or\ 3\Phi=-\dfrac{\pi}{3}+2k\pi\qquad\text{divide both sides by 3}\\\\\Phi=\dfrac{\pi}{9}+\dfrac{2k\pi}{3}\ or\ \Phi=-\dfrac{\pi}{9}+\dfrac{2k\pi}{3}[/tex]
