I am really stuck on this problem and don't even know where to begin.
The problem: The equation $$3\sin\bigl(k(x−10)\bigr) = 1$$ has 3 solutions on the interval $30°\le x\le 180°$. Determine/calculate all possible positive values of the constant k.
I have been working on this forever, but I really don't get it and any kind of help would really be appreciated.
Let $\theta = \arcsin \frac 13$ and let $u = x - 10$. Then the solutions are $$ku = \theta + 360n\\ku = 180 - \theta +360n$$ or $$u = \frac {\theta + 360n}k\\u = \frac {180 - \theta + 360n}k$$ and you are looking for exactly three solutions of $u$ between $40^\circ$ and $190^\circ$.
What is most helpful at this point is to look at the distance between these solutions. That distance grows or shrinks with $k$. You need to fit exactly 3 solutions in a range that is $150^\circ$ wide. When $k$ is too small, the distance will grow enough that any 3 adjacent solutions will cover a range of more than $150$ degrees. When $k$ is too large, any range of $150$ is going to contain 4 or more solutions.
So start by finding the smallest $k$ for which it is possible to fit three solutions in an 150 degree range. And then the smallest $k$ for which it is guaranteed to get 4 solutions with any 150 degree range. Since this doesn't consider the specific 150 degree range you are interested in, the values of $k$ must fall in this range, but need not be these particular values.
Once you have that figured out, then you can consider what actually happens in that when you turn the $k$-knob up from the minimum to the maximum. When $k$ is at the bottom, you will have two values of $u$ between $40$ and $190$, with the next values outside in each direction. As $k$ increases, all of the solutions start moving clockwise on the circle towards $0^\circ$. At some point, the third solution will reach $190$ and enter your range. That is the first value you seek, as you keep increasing $k$, a 4th solution will approach the $190$ limit - but at the same time, the lowest value in the interval is also approaching $40$. You will need to find both when the 4th solution reaches $190$, and where the 1st solution reaches $40$ and exits the range. If the 4th solution enters before the 1st solution exits, then you will have 4 solutions in the range, otherwise you will have only 2 solutions. In either case, you've entered a region where the values of $k$ are not allowed.
Keep increasing $k$ until the next solution enters. Now you are back to 3 solutions and allowable values of $k$. Keep going like this until you reach the upper limit where 4 solutions are guaranteed to be in the range, no matter how it is positioned. At that point, you know you have identified all values of $k$.