Chapter 7 - Trigonometry and Periodic Functions - 7.8 Inverse Trigonometric Functions - Exercises and Problems for Section 7.8 - Exercises and Problems - Page 314: 21

$$t=\frac{\pi }{3},\:t=\frac{4\pi }{3}$$

Using inverse trigonometric functions to solve the equation, we find: $$\tan \left(t\right)=\sqrt{3},\:0\le \:t\le \:2\pi \\ t=\frac{\pi }{3}+\pi n \\ t=\frac{\pi }{3},\:t=\frac{4\pi }{3}$$