$\cos2\theta +\cos\theta +k = 0 $ - set of all values of $k$ for which there is a solution 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?
 A: 
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 $\cos\theta=1$.
A: HINT:
$$2\cos^2\theta+\cos\theta+k-1=0\iff(4\cos\theta+1)^2=9-8k$$
As $-1\le\cos\theta\le1,$ 
$-3\le4\cos\theta+1\le5\implies0\le(4\cos\theta+1)^2\le25$
A: $$\cos 2\theta +\cos \theta=-k$$
$$2\cos^ 2\theta+\cos \theta-1=-k$$
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 $$-2\le k\le \frac98$$
A: $2\cos^2\theta -1+\cos\theta+k=0$  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 $\theta=0$
