<s>[INST] Find the values of $(2\sin x-1)(\cos x+1)=0$ How many different value of x from 0° to 180° for the equation $(2\sin x-1)(\cos x+1) = 0$?
The solution shows that one of these is true:
$\sin x = \frac12$ and thus $x = 30^\circ$ or $120^\circ$ 
$\cos x = -1$ and thus $x = 180^\circ$
Question: Inserting the $\arcsin$ of $1/2$ will yield to $30°$, how do I get $120^\circ$? and what is that $120^\circ$, why is there $2$ value but when you substitute $\frac12$ as $x$, you'll only get $1$ value which is the $30^\circ$?
Also, when I do it inversely: $\sin(30^\circ)$ will result to 1/2 which is true as $\arcsin$ of $1/2$ is $30^\circ$. But when you do $\sin(120^\circ)$, it will be $\frac{\sqrt{3}}{2}$, and when you calculate the $\arcsin$ of $\frac{\sqrt{3}}{2}$, it will result to $60^\circ$ and not $120^\circ$. Why?
 [/INST] If $ab = 0$, then either $a=0$ or $b=0$.
Given $(2\sin x - 1)(\cos x + 1) = 0$, we can separate these and solve separately.
So... $\ \ \ \ \ 2\sin x -1 = 0$
$\Rightarrow \sin x = \frac12 \Rightarrow x = 30^\circ, 150^\circ$
We get those two values because $\sin x$ is symmetrical across the $y$-axis.
Also... $\ \ \ \ \ \cos x + 1 = 0$
$\Rightarrow \cos x = -1 \Rightarrow x = 180^\circ$
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