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?
Best Answer
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.