## Trigonometry (11th Edition) Clone

The solution set to this problem is $$\{0^\circ,45^\circ, 135^\circ,180^\circ, 225^\circ, 315^\circ\}$$
$$\sec^2\theta\tan\theta=2\tan\theta$$ over interval $[0^\circ,360^\circ)$ 1) Solve the equation: $$\sec^2\theta\tan\theta=2\tan\theta$$ $$\sec^2\theta\tan\theta-2\tan\theta=0$$ $$\tan\theta(\sec^2\theta-2)=0$$ $$\tan\theta=0\hspace{1cm}\text{or}\hspace{1cm}\sec^2\theta=2$$ $$\tan\theta=0\hspace{1cm}\text{or}\hspace{1cm}\sec\theta=\pm\sqrt2$$ 2) Apply the inverse function: - For $\tan\theta=0$: Over the interval $[0^\circ,360^\circ)$, there are 2 values of $\theta$ where $\tan\theta=0$, which are $\{0^\circ,180^\circ\}$ - For $\sec\theta=\sqrt2$: Since $\sec\theta=\frac{1}{\cos\theta}$, having $\sec\theta=\sqrt2$ means $\cos\theta=\frac{\sqrt2}{2}$. Over the interval $[0^\circ,360^\circ)$, there are 2 values of $\theta$ where $\cos\theta=\frac{\sqrt2}{2}$, which are $45^\circ$ in quadrant I and $315^\circ$ in quadrant IV. - For $\sec\theta=-\sqrt2$: Since $\sec\theta=\frac{1}{\cos\theta}$, having $\sec\theta=-\sqrt2$ means $\cos\theta=-\frac{\sqrt2}{2}$. Over the interval $[0^\circ,360^\circ)$, there are 2 values of $\theta$ where $\cos\theta=-\frac{\sqrt2}{2}$, which are $135^\circ$ in quadrant II and $225^\circ$ in quadrant III. Therefore, $$\theta\in\{0^\circ,45^\circ, 135^\circ,180^\circ, 225^\circ, 315^\circ\}$$ In other words, the solution set to this problem is $$\{0^\circ,45^\circ, 135^\circ,180^\circ, 225^\circ, 315^\circ\}$$