Answer
The solution set to this problem is $$\{106.3^\circ, 149.6^\circ,286.3^\circ,329.6^\circ\}$$
Work Step by Step
$$\tan^2\theta+4\tan\theta+2=0$$ 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 $$\{106.3^\circ, 149.6^\circ,286.3^\circ,329.6^\circ\}$$
