This looks like a trigonometric equation involving sine and cosine of x. I know that $\sin^2 x = 1 - \cos^2 x,$ so I can use that identity to replace $\sin^2 x$ with $1 - \cos^2 x$ in the equation. This gives me $1 - \cos^2 x + \cos x + 1 = 0.$ Now I have a quadratic equation in terms of $\cos x.$ I can rewrite it as $\cos^2 x - \cos x - 2 = 0.$ To solve for $\cos x,$ I can try to factor the quadratic expression. I need two numbers that multiply to -2 and add to -1. I can see that -2 and 1 work. So I can write $\cos^2 x - \cos x - 2 = (\cos x - 2)(\cos x + 1) = 0.$ Now I can use the zero product property to say that either $\cos x - 2 = 0$ or $\cos x + 1 = 0.$ Solving for $\cos x,$ I get $\cos x = 2$ or $\cos x = -1.$ However, I know that $\cos x = 2$ is impossible, since the cosine function only takes values between -1 and 1. So the only possible value for $\cos x$ is -1. This means that x must be an angle whose cosine is -1. I know that $\cos \pi = -1,$ so that is one solution. I also know that $\cos x$ is periodic with period $2 \pi,$ so any angle that is a multiple of $2 \pi$ away from $\pi$ will also have a cosine of -1. Therefore, the general solution is $x = \pi + 2 \pi k,$ where k is any integer. However, the problem asks for solutions in the interval $[0, 2 \pi].$ This means that I only need to consider values of k that make $x$ lie in that interval. I can see that when k = 0, I get $x = \pi,$ which is in the interval. When k = 1, I get $x = 3 \pi,$ which is not in the interval. When k = -1, I get $x = -\pi,$ which is also not in the interval. Therefore, the only solution in the given interval is $x = \pi.$ 
Find all values of $x,$ $0 \le x \le 2 \pi,$ which satisfy
\[\sin^2 x + \cos x + 1 = 0.\]Enter all the solutions, separated by commas.