Answer
The solution set to this problem is $$\{\varnothing\}$$
Work Step by Step
$$\sin^2\theta-2\sin\theta+3=0$$ over interval $[0^\circ,360^\circ)$
1) Solve the equation:
$$\sin^2\theta-2\sin\theta+3=0$$
Here we need to consider the equation a quadratic formula. In fact, if we take $\sin\theta$ as $x$, we would get
$$x^2-2x+3=0$$
with $a=1, b=-2$ and $c=3$
- Calculate $\Delta$: $\Delta=b^2-4ac=(-2)^2-4\times1\times3=4-12=-8\lt0$
$\Delta\lt0$ 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 $$\sin^2\theta-2\sin\theta+3=0$$
Therefore, the solution set to this problem is $$\{\varnothing\}$$
