For which value of $k$ the trigonometric function $\sin x + \cos^2(x) +k = 0 $ doesn't have solutions? 
For which values of $k$ does the trigonometric equation $\sin x + \cos^2(x) + k =0$ have no solution?

Rearranging the solutions are from the equation :
$  -\sin^2(x) + \sin(x) + 1 + k = 0$
By the fact that it is a quadratic equation in $\sin(x) $, by imposing the discriminant to be equal to zero, it should give us the range of values of k, which is:
$k < -5/4$
however, by drawing the graph of the function on Geogebra, it looks like there is another range, which I couldn't find algebraically, which is:
$k > 1 $
so that the final solutions should be:
$k < -5/4 ∨ k > 1$
THE QUESTION IS: how can i find the latter solution without looking at the graph, but simply using algebra?
 A: We complete the square :
$-\sin^2(x) + \sin(x) + 1 + k = -(\sin(x)-1/2)^2 + 1/4 +1 + k$
$\sin(x)$ can take all values on $[-1,1]$
This lets the negative square take values $-(-3/2)^2= -9/4$ to $-(0)^2 = 0$
So we get 2 different values for $k$ $$-9/4+5/4+k = 0   \Leftrightarrow k = 1$$
$$0+5/4+k = 0   \Leftrightarrow k = -5/4$$
Now since we only found the end points we must additionally figure out if it is allowed between $-5/4$ and $1$ or outside of this interval.
A: For the equation $\sin^2x -\sin x-(1+k)=0$ to have a solution we require
(1) the discriminant $D:=5+4k$ must be non-negative, and
(2) at least one of the two roots $\frac{1\pm\sqrt D}2$ must lie between $-1$ and $1$.
Condition (1) means $k\ge -\frac54$.
Condition (2) breaks into two requirements, at least one must be satisfied by $D$:
(2a) $-1\le\frac{1 +\sqrt D}2 \le 1$, which is the same as $\sqrt D\le 1$.
(2b) $-1\le\frac{1 -\sqrt D}2 \le 1$, which is the same as $\sqrt D\le 3$.
The requirement "$\sqrt D\le1$ or $\sqrt D\le3$" is the same as just $\sqrt D\le 3$. So the additional requirement for the equation to have a solution is $\sqrt D\le 3$, which means $k\le1$.
Flipping this around, no solution exists when $k<-\frac54$ or $k>1$.
