\[ 2 \cos ^{2} x+\cos x-1=0 \] A. The solution sol is $\left\{\frac{\pi}{2}+2 h t, \frac{5 t}{8}+2 t i t,+2 h t\right.$, where it is any integer $\}$ B. The solution set is $\left\{\frac{\pi}{2} * 2 n t, \frac{5 t}{3}+2 n t, t+2 n t\right.$, where $n$ is any integer $\}$ C. The solution sel is $\left\{\frac{\pi}{3}+2 n \pi, \frac{5 \pi}{3}+2 n t, x+2 n t\right.$, where $n$ is any integer $\}$. D. The solution sot is $\left\{\frac{\pi}{3}+2 n \pi, \frac{5 \pi}{8}+2 n \pi, \pi+2 n \pi\right.$, where $n$ is any integer $\}$

Step 1 ：The given equation is a quadratic equation in terms of \(\cos x\). We can solve it by using the quadratic formula. The solutions of the equation will give us the values of \(\cos x\). We can then use the properties of the cosine function to find the general solutions for \(x\).

Step 2 ：The solutions for \(x\) are \(\pi\) and \(\pi/3\). However, these are not the general solutions. The general solutions can be obtained by adding \(2n\pi\) to each solution, where \(n\) is any integer. This is because the cosine function has a period of \(2\pi\).

Step 3 ：The general solutions for \(x\) are \(\boxed{\left\{2\pi n + \pi, 2\pi n + \frac{\pi}{3}\right\}}\), where \(n\) is any integer. Therefore, the correct option is D. The solution set is \(\left\{\frac{\pi}{3}+2 n \pi, \pi+2 n \pi\right\}\), where \(n\) is any integer.