Chapter 9 - Trigonometric Identities, Models, and Complex Numbers - 9.1 Trigonometric Equations - Exercises and Problems for Section 9.1 - Exercises and Problems - Page 357: 7

$$θ=\frac{\pi }{20},\:θ=\frac{\pi }{4},\:θ=\frac{9\pi }{20}$$

Solving the equation using inverse trigonometric functions, we find: $$9\tan \left(5θ\right)+1=10,\:0\le \:θ\le \frac{\pi }{2} \\ \tan \left(5θ\right)=1 \\ θ=\frac{\pi }{20}+\frac{\pi n}{5} \\ θ=\frac{\pi }{20},\:θ=\frac{\pi }{4},\:θ=\frac{9\pi }{20}$$