Find all solutions of the equation in the interval $[0,2 \pi)$.
\[
2 \cos \theta+\sqrt{3}=0
\]
Write your answer in radians in terms of $\pi$.
If there is more than one solution, separate them with commas.

Step 1 ：First, we need to isolate the cosine function. We can do this by subtracting \(\sqrt{3}\) from both sides of the equation. This gives us \(2 \cos \theta = -\sqrt{3}\).

Step 2 ：Next, we divide both sides of the equation by 2 to solve for \(\cos \theta\). This gives us \(\cos \theta = -\frac{\sqrt{3}}{2}\).

Step 3 ：We know that \(\cos \theta = -\frac{\sqrt{3}}{2}\) when \(\theta = \frac{5\pi}{6}\) or \(\theta = \frac{7\pi}{6}\) in the interval \([0, 2\pi)\).

Step 4 ：Therefore, the solutions to the equation \(2 \cos \theta+\sqrt{3}=0\) in the interval \([0,2 \pi)\) are \(\theta = \frac{5\pi}{6}, \frac{7\pi}{6}\).

Step 5 ：Finally, we check our solutions. Substituting \(\theta = \frac{5\pi}{6}\) into the original equation, we get \(2 \cos \frac{5\pi}{6}+\sqrt{3}=0\), which simplifies to \(0=0\). Substituting \(\theta = \frac{7\pi}{6}\) into the original equation, we get \(2 \cos \frac{7\pi}{6}+\sqrt{3}=0\), which also simplifies to \(0=0\). Therefore, our solutions are correct.

Step 6 ：So, the solutions to the equation \(2 \cos \theta+\sqrt{3}=0\) in the interval \([0,2 \pi)\) are \(\boxed{\theta = \frac{5\pi}{6}, \frac{7\pi}{6}}\).