Answer
The solution set is $$\{33.6^\circ+360^\circ n, 326.4^\circ+360^\circ n,n∈Z\}$$
Work Step by Step
$$5+5\tan^2\theta=6\sec\theta$$
1) Solve the equation over the interval $[0^\circ,360^\circ)$
$$5+5\tan^2\theta=6\sec\theta$$
$$5(1+\tan^2\theta)=6\sec\theta$$
- Recall the identity: $1+\tan^2\theta=\sec^2\theta$
$$5\sec^2\theta=6\sec\theta$$
$$5\sec^2\theta-6\sec\theta=0$$
$$\sec\theta(5\sec\theta-6)=0$$
$$\sec\theta=0\hspace{1cm}\text{or}\hspace{1cm}\sec\theta=\frac{6}{5}$$
- For $\sec\theta=0$
The range of a sectant function is $(−\infty,−1]∪[1,\infty)$. Since $0\notin(−\infty,−1]∪[1,\infty)$, there cannot be any values of $\theta\in[0^\circ,360^\circ)$ that have $\sec\theta=0$.
- For $\sec\theta=\frac{6}{5}$. That means, $$\theta=\sec^{-1}\frac{6}{5}$$
To find $\sec^{-1}\frac{6}{5}$ when even calculator does not have that function, recall that $\sec\theta=\frac{1}{\cos\theta}$.
So, $\sec^{-1}x=\cos^{-1}\Big(\frac{1}{x}\Big)$
Therefore, $$\theta=\sec^{-1}\frac{6}{5}=\cos^{-1}\frac{5}{6}$$
$$\theta\approx33.6^\circ\hspace{1cm}\text{or}\hspace{1cm}\theta\approx326.4^\circ$$
Therefore, overall, $$\theta=\{33.6^\circ,326.4^\circ\}$$
2) Solve the equation for all solutions
- The integer multiples of the period of the secant function is $360^\circ$.
- We apply it to each solution found in step 1, meaning the solution set would be
$$\{33.6^\circ+360^\circ n, 326.4^\circ+360^\circ n,n∈Z\}$$
