## Functions Modeling Change: A Preparation for Calculus, 5th Edition

$$θ=\frac{\pi }{12}$$
Solving the equation using inverse trigonometric functions, we find: $$2\sqrt{3}\tan \left(2θ\right)+1=3,\:0\le \:θ\le \frac{\pi }{2} \\ \tan \left(2θ\right)=\frac{\sqrt{3}}{3} \\ θ=\frac{\pi }{12}+\frac{\pi n}{2} \\ θ=\frac{\pi }{12}$$