Chapter 9 - Trigonometric Identities, Models, and Complex Numbers - 9.1 Trigonometric Equations - Exercises and Problems for Section 9.1 - Exercises and Problems - Page 357: 19

$$ x=1.107 ,\:x=2\pi -1.107,\:x=2.03 ,\:x=-2.03 +2\pi $$

Solving the equation using inverse trigonometric functions, we find: $$5\left(\cos ^2\left(x\right)\right)=1,\:0\le \:x\le \:2\pi \\ \cos \left(x\right)=\sqrt{\frac{1}{5}},\:\cos \left(x\right)=-\sqrt{\frac{1}{5}} \\ x=1.107 ,\:x=2\pi -1.107,\:x=2.03 ,\:x=-2.03 +2\pi $$