Solving trigonometric equation $\sin(2x) = \frac{1}{2}$ I have tried to solve trigonometric equations in $2$ different ways. I know that only the Case $1$ is the correct way of solving the equation, but I don’t understand why case $2$ is an incorrect way of solving the equation.
After looking at the respective unit circle of case $1$ and $2$ I feel like case $2$ should be correct since both the answers of case $2, 15, 165$ have the same $y$ coordinates on the unit circle i.e. they have the same sin theta value. Whereas the values of case $1$ have different $y$ coordinates and hence different value of $\sin (\theta)$.
Can someone explain why case $2$ is incorrect by using the unit circle in the explanation.
Case $1$:
$\sin (2x)$ = $\frac{1}{2}$
$\arcsin (\frac{1}{2}) = 2x$
$30 = 2x$
$180 - 30 = 2x $
$150 = 2x $
$X = 15, 75$

Case $2$:
$\sin (2x) = \frac{1}{2}$
$\arcsin (\frac{1}{2}) = 2x$
$30 = 2x$
$X = 15$
$X = 180 - 15$
$X = 165$

 A: I'd say they are both wrong.
The both make the mistake of claiming

$\sin W = k\implies W = \arcsin k$.

This is simply not true.  For example $\sin 150^\circ = \frac 12$ but it is most certainly not the case that $150^\circ = 30^\circ = \arcsin \frac 12$.  What they both should be saying is: If $\sin W=k$ then $W=\arcsin k$ is one possible solution.  And $W = 180- \arcsin k$ is another possible value.  And $W= \arcsin k \pm 360m$ or $W=180 -\arcsin k \pm 360n$ for any integers $m,n$ are an infinite number of possible solutions.
$\arcsin k$ is only just one value and it is always a value between $-90$ and $90$.  It is not two or more values at the same time and it is never all the solutions to $\sin W=k$.  It is not that $\arcsin \frac 12 = 30, 150$.  It's that $\arcsin \frac 12 = 30$ only.  And is that the solutions so $\sin W = \frac 12$ include $W= \arcsin \frac 12 =30$ as one solution and $W =180- \arcsin \frac 12=180-30 =150$ as another solutions.
Then after they made this error and claimed that $W=\arcsin k = M$, they attempt to fix the error be making the claim

If $W = M$ then $W = 180-M$.

By itself this is just plain nuts.  After only  $27 = 27$ but $27 \ne 180-27 = 153$.
But the attempt is to allow for two possible solutions and to make up for the mistake above.
By making up for $W= 30$ (i.e. $W= \arcsin \frac 12 = 30$) then they are allowing for $W= 180 -30$ (i.e. $W=180 -\arcsin \frac 12 = 30$) as another solution and that is why, even though it is wrong, the first method gets the correct answers.
Method 1 says if $W = 2x =30$ or $W= 2x = 150$ then $x = 15$ or $75$.
Method 2 always took one solution $2x=30$ and divided by $2$ to get $x = 15$.  At this point $x = \frac W2$ is not the angle that was entered into $\sin$ expression.  But method 2 attempt to say if $x = 15$ then $x =180-15$ is still okay.
It's not.  They only reason method 1 got a wrong answer by claiming $2x =W = 30$ so $2x = W = 180-30$ got a right answer was because $2x = W = \arcsin \frac 12;2x = W = 180 -\arcsin \frac 12$ are the two valid possible solutions.
But if we do $2x = W=30$ and $x=\frac W2 = 15$ we no longer have $x =\arcsin SOMETHING$.  And whatever logistics there was for subtracting $W=\arcsin \frac 12$ from $180$ there is utterly no reason to subtract $x \ne \arcsin ANYTHING$ from $180$.
tl;dr
I'd say the correct way to do it is as follows.
$\sin 2x = \frac 12$.  If we are considering the two values between $-90$ and $270$ then they would be:
$2x =\arcsin \frac 12$ OR $2x = 180 -\arcsin \frac 12$.
As $\arcsin \frac 12 = 30$ then
$2x = 30$ OR $2x=180 -30=150$.
So $x = 15$ or $x = 75$.
A: Note the only difference is that in the first, 180 is being used to find 2x, while in the second, 180 is to find x.
When working with sin(2x), you want to multiply the degrees by 2x. The reason subtracting from 180 works is because you're still working with 2x, but when you subtract 15 from 180 the 180 is inserted for x.
