# 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?

• by solutions do you mean zeros?
– Mark
Sep 28, 2021 at 20:26
• Letting $y=\sin x$, you get a trinomial in $y$, and you can find when it has real solutions, but it's not enough: the values of $y$ must be in $[-1,1]$. For instance it's quite obvious from the equation that if $k>2$, there can be no solution. You have thus to solve an inequation, or rather find when it has solutions. Sep 28, 2021 at 20:27
• What you have is not an equation. Hint: an EQUATion must have an EQUAL sign in it. Sep 28, 2021 at 20:33
• @ arsenio: Please feel free to roll back if my edit not ok. Sep 28, 2021 at 22:46
• The subject line and the quoted problem are different. One is set equal to $y$, the other to $0$. The one in the subject line always has solutions, because for any particular value of $k$ and $x$, one can simply set $y$ to be the appropriate value to get an equality. In short: this is still a total mess. Please clean it up. Sep 29, 2021 at 3:20

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.

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 $$k>1$$.