Find $m$ so that the equation has real solution:
$$\cos^2 x + (m+1) \sin x = 2m - 1$$
This can be rewritten as
$$2 \sin^2 x - (m+1) \sin x +2m - 2 = 0$$
A first condition would be the discriminant $\Delta \ge 0$, from which I got that $m \in (-\infty, 7-4 \sqrt{2}] \cup [7 + 4 \sqrt{2}, + \infty)$.
Also, the minimum point of the function should be in $[-1, 1]$, so another condition would be $\frac{m+1}{2} \in [-1, 1]$. What else should I impose ?
The correct answer is $ m \in [0, 2]$.
you must solve the equation $$\sin(x)^2-(m+1)\sin(x)+2m-2=0$$ we have exactly two real Solutions namely $2$ or $$m-1$$ can you finish? and we have $$-1\le m-1\le 1$$