Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.2 Trigonometric Equations I - 6.2 Exercises - Page 267: 56

The solution set is $$\{71.6^\circ+180^\circ n, 135^\circ+180^\circ n, n\in Z\}$$

$$\sec^2\theta=2\tan\theta+4$$ 1) Solve the equation over the interval $[0^\circ,360^\circ)$ $$\sec^2\theta=2\tan\theta+4$$ $$\sec^2\theta-2\tan\theta-4=0$$ - Recall the identity: $1+\tan^2\theta=\sec^2\theta$ $$1+\tan^2\theta-2\tan\theta-4=0$$ $$\tan^2\theta-2\tan\theta-3=0$$ $$(\tan^2\theta+\tan\theta)+(-3\tan\theta-3)=0$$ $$\tan\theta(\tan\theta+1)-3(\tan\theta+1)=0$$ $$(\tan\theta+1)(\tan\theta-3)=0$$ $$\tan\theta=-1\hspace{1cm}\text{or}\hspace{1cm}\tan\theta=3$$ - For $\tan\theta=-1$ Over the interval $[0^\circ, 360^\circ)$, there are 2 values of $\theta$ where $\tan\theta=-1$, which are $135^\circ$ and $315^\circ$. So, $\theta=\{135^\circ, 315^\circ\}$ - For $\tan\theta=3$. That means, $$\theta=\tan^{-1}3$$ $$\theta\approx71.6^\circ\hspace{1cm}\text{or}\hspace{1cm}\theta\approx251.6^\circ$$ Therefore, overall, $$\theta=\{71.6^\circ,135^\circ, 251.6^\circ, 315^\circ\}$$ 2) Solve the equation for all solutions - The integer multiples of the period of the tangent function is $180^\circ$. - We apply it to each solution found in step 1. - We end up with a set like this $$\{71.6^\circ+180^\circ n, 135^\circ+180^\circ n, 251.6^\circ+180^\circ n, 315^\circ+180^\circ n,n∈Z\}$$ - However, as $251.6^\circ+180^\circ n$ and $71.6^\circ+180^\circ n$ refer to the same set of values, we only need to include one of them in the final solution set. The same thing can be said about $135^\circ+180^\circ n$ and $315^\circ+180^\circ n$. So the solution set is $$\{71.6^\circ+180^\circ n, 135^\circ+180^\circ n, n\in Z\}$$