Answer
The solution set to this problem is $$\{\varnothing\}$$
Work Step by Step
$$2\cos^2\theta+2\cos\theta+1=0$$ over interval $[0^\circ,360^\circ)$
1) Solve the equation:
$$2\cos^2\theta+2\cos\theta+1=0$$
Here we need to consider the equation a quadratic formula. In fact, if we take $\cos\theta$ as $x$, we would get
$$2x^2+2x+1=0$$
with $a=2, b=2$ and $c=1$
- Calculate $\Delta$: $\Delta=b^2-4ac=2^2-4\times2\times1=4-8=-4\lt0$
$\Delta\lt0$ means that there would be no values of $x\in R$ which can satisfy the given equation.
That also means there would be no values of $\theta\in[0^\circ,360^\circ)$ which can satisfy the original equation $$2\cos^2\theta+2\cos\theta+1=0$$
Therefore, the solution set to this problem is $$\{\varnothing\}$$
