$\cos2\theta +\cos\theta +k = 0 $ - set of all values of $k$ for which there is a solution

Question

The set of all values of $k$ (real), such that the equation $\cos2\theta +\cos\theta +k = 0 $ admits a solution for $\theta$ is?
MY ATTEMPT:
I substituted $\cos2\theta$ with $2\cos^2\theta - 1 $. On completing the square in the given I found out that the minimum value of k is -2. But that is wrong. It should be $0$. Right? So how should I solve? Also how do I find the maximum as well?

Answer 1

HINT:

$$2\cos^2\theta+\cos\theta+k-1=0\iff(4\cos\theta+1)^2=9-8k$$

As $-1\le\cos\theta\le1,$

$-3\le4\cos\theta+1\le5\implies0\le(4\cos\theta+1)^2\le25$

Answer 2

It should be 0.Right?

No, the minimum value of $k$ is not $0$.

One way is to represent $k$ by $\cos\theta$. We have $$k=-2\cos^2\theta-\cos\theta+1=-2\left(\cos\theta+\frac 14\right)^2+\frac 98$$ Now, considering the parabola with $-1\le\cos\theta\le 1$ gives that $\color{red}{k_{\text{max}}=\frac 98}$ when $\cos\theta=-1/4$, and $\color{red}{k_{\min}=-2}$ when $\cos\theta=1$.

Answer 3

$$\cos 2\theta +\cos \theta=-k$$ $$2\cos^ 2\theta+\cos \theta-1=-k$$ Let $\cos \theta=t$, $-1\le t \le1$ and $f(t)=2t^2+t-1$

If $-1\le t \le1$, then $$-\frac98\le f(t)\le2$$

Then $$-\frac98\le -k\le2$$ Hence $$-2\le k\le \frac98$$

Answer 4

$2\cos^2\theta -1+\cos\theta+k=0$ let $\cos\theta=x$ then $-k=2x^2+x-1=p(x)$ (let)
$-1\le\cos\theta\le1$ or $-1 \le x \le 1$

$p(-1)=0$ and $p(1)=2$ now differentiation of $p(x)=0$ give us the minima so $x=-1/4$ is minima
$p(-1/4)=-9/8$
so $-9/8 \le p(x) \le 2$
so $-9/8 \le -k \le 2$
so $9/8 \ge k \ge -2$

Now your question it can be $-2$ put $\theta=0$