Solve the equation for the interval $[0,2 \pi)$.
\[
\cos ^{2} x+2 \cos x+1=0
\]

Step 1 ：We are given the equation \(\cos ^{2} x+2 \cos x+1=0\) and asked to solve it for the interval \([0,2 \pi)\).

Step 2 ：This equation is a quadratic equation in terms of \(\cos x\). We can solve it by using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 1\), \(b = 2\), and \(c = 1\).

Step 3 ：Substituting these values into the quadratic formula, we find that the roots of the quadratic equation are both -1.

Step 4 ：These roots correspond to an angle of \(\pi\) in the interval \([0,2 \pi)\).

Step 5 ：However, the cosine function is periodic with a period of \(2 \pi\), so we need to check if there are other solutions in the interval \([0,2 \pi)\).

Step 6 ：Upon checking, we find that the other possible solutions are also \(\pi\).

Step 7 ：Therefore, the only solution to the equation in the interval \([0,2 \pi)\) is \(\pi\).

Step 8 ：\(\boxed{\pi}\) is the final answer.