## Trigonometry (11th Edition) Clone

The solution set to this problem is $$\{106.3^\circ, 149.6^\circ,286.3^\circ,329.6^\circ\}$$
$$\tan^2\theta+4\tan\theta+2=0$$ over interval $[0^\circ,360^\circ)$ 1) Solve the equation: $$\tan^2\theta+4\tan\theta+2=0$$ Here we need to consider the equation a quadratic formula. In fact, if we take $tan\theta$ as $x$, we would have $$x^2+4x+2=0$$ with $a=1, b=4$ and $c=2$ - Calculate $\Delta$: $\Delta=b^2-4ac=4^2-4\times1\times2=16-8=8$ - Calculate $x$, or in other words, $\tan\theta$: $$x=\tan\theta=\frac{-b\pm\sqrt\Delta}{2a}=\frac{-4\pm\sqrt{8}}{2}=\frac{-4\pm2\sqrt2}{2}=-2\pm\sqrt2$$ 2) Apply the inverse function: - For $\tan\theta=-2+\sqrt2$: $$\tan\theta=-2+\sqrt2\approx-0.586$$ which means $$\theta=\tan^{-1}(-0.586)$$ $$\theta\approx149.6^\circ\hspace{1cm}\text{or}\hspace{1cm}\theta\approx329.6^\circ$$ - For $\tan\theta=-2-\sqrt2$: $$\tan\theta=-2-\sqrt2\approx-3.414$$ which means $$\theta=\tan^{-1}(-3.414)$$ $$\theta\approx106.3^\circ\hspace{1cm}\text{or}\hspace{1cm}\theta\approx286.3^\circ$$ Therefore, $$\theta\in\{106.3^\circ, 149.6^\circ,286.3^\circ,329.6^\circ\}$$ In other words, the solution set to this problem is $$\{106.3^\circ, 149.6^\circ,286.3^\circ,329.6^\circ\}$$