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?

  • $\begingroup$ by solutions do you mean zeros? $\endgroup$
    – Mark
    Sep 28, 2021 at 20:26
  • $\begingroup$ 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. $\endgroup$ Sep 28, 2021 at 20:27
  • 2
    $\begingroup$ What you have is not an equation. Hint: an EQUATion must have an EQUAL sign in it. $\endgroup$ Sep 28, 2021 at 20:33
  • $\begingroup$ @ arsenio: Please feel free to roll back if my edit not ok. $\endgroup$
    – Narasimham
    Sep 28, 2021 at 22:46
  • $\begingroup$ 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. $\endgroup$ Sep 29, 2021 at 3:20

2 Answers 2


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 $k>1$.


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