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 $2$:
$\sin (2x) = \frac{1}{2}$
$\arcsin (\frac{1}{2}) = 2x$
$30 = 2x$
$X = 15$
$X = 180 – 15$
$X = 165$
Best Answer
I'd say they are both wrong.
The both make the mistake of claiming
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
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 $x = 75$.