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?
 A: As you noted $x=120°$ is not a solution indeed


*

*$(2\sin 120°-1)(\cos 120°+1)=\left(2\frac{\sqrt3}2-1\right)\left(-\frac12+1\right)\neq 0$


but


*

*$2\sin x-1=0 \implies x=\frac16 \pi + 2k\pi,\,x=\frac56 \pi + 2k\pi$

*$\cos x+1 = 0 \implies x=\pi + 2k\pi=(2k+1)\pi$

A: The function $\arcsin$ returns the angle within the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$. In fact, there are infinitely many angles whose sine is $1/2$. Since a function can't spit out more than one number however, we define the $\arcsin$ function to return the inverse of the sine on that interval.
Anyways, the true answer here is actually $30$ degrees and $150$ degrees (with also $180$ from the cosine expression). How would you find this? Just look at our unit circle.

The part we're restricted to is the top half of the circle going counterclockwise (0 to 180). What values of $x$ spit out a y-coordinate (sine) of $1/2$?  Well, $\pi/6$ and $5\pi/6$ (or 30 and 150). There's your answer!
A: 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... $\ \ \ \ \ \cos x + 1 = 0$
$\Rightarrow \cos x = -1 \Rightarrow x = 180^\circ$
A: $0\leq \sin(x)\leq1$ in quadrants I and II. That is to say $\sin x$ (and $\cos x$) are stuck in a circle which is periodic. There are an infinite amount of $x$ that satisfy for example $\sin(x)=\frac{1}{2}$, but the simplest answer is enough to suffice
Specifically, $\sin^{-1}x$ isn't a function unless its domain is restricted to $-\pi/2\leq x\leq\pi/2$
Likewise for $\cos^{-1}x$. Its not a function unless its domain is restricted to $0\leq x\leq\pi$
