Well that's the question, I have the graph of the function in the interval $0 \lt x \lt \frac{3 \pi}{2 }$and I need to know (in order to get a feasible set) when this inequality stands:$ y- \sin(x) \ge 0$, I tried cutting the function in $\pi$, and then I gave two values in both sides of areas, $\gt \pi$ and $\lt \pi$, but the function is negative when $x \lt \pi $ even when the graph shows the area over the $x$ axis... so I don't understand... I also don't understand where should the feasible region be... I mean, should I paint all the area until reach $\pi$ assuming that only one side of the total area is feasible?
Sorry for this silly question, it's just that recently I had to take a lot of this again soIi can get the basics for non linear optimization.
I would really appreciate your help. Thanks
Note that when written in standard form, this equation is $y \geq \sin{x}$. This means that all values of y in the entire coordinate plane that are greater than or equal to $\sin{x}$ will yield a valid solution.
The thing with the question they gave you is that it does not ask for a particular set of x such that $\mathbf{\sin{x}\geq0}$. The answer to the previous question is $[0,\pi]$ because $\sin{x}\geq0$ in this region. This is NOT the question they asked you.
What they asked you was to find all points (x,y) that satisfy the inequality- including points that are not part of the graph $y \mathbf{=}\sin{x}$ (Notice the equals sign). Take a look at the graph https://www.desmos.com/calculator/fhgwbtbgdu (Notice that the curve $\sin{x}$ of the graph is a solid line instead of a dotted because it is $\geq$ instead of $\gt$.
Your inequality is as simplified as it can get, so technically that is your answer or you can represent your answer through the graph. That region is all the points where the solution is valid. For example, take $(2,\pi)$ as a point. A quick check will show you that $$y-\sin{x}\geq0$$$$2-\sin{\pi}\geq0$$$$2-0\geq0$$$$2\geq0$$ which is true.