Chapter 7 - Trigonometry and Periodic Functions - 7.8 Inverse Trigonometric Functions - Exercises and Problems for Section 7.8 - Exercises and Problems - Page 314: 16

$$t=\frac{\pi }{6},\:t=\frac{5\pi }{6}$$

Using inverse trigonometric functions to solve the equation, we find: $$\sin \left(t\right)=\frac{1}{2},\:0\le \:t\le \:2\pi \\ t=\frac{\pi }{6}+2\pi n,\:t=\frac{5\pi }{6}+2\pi n \\ t=\frac{\pi }{6},\:t=\frac{5\pi }{6}$$ NOTE: All answers are in radians.