Solve the trigonometric equation for all values \( 0 \leq x<2 \pi \) \[ 2 \cos x=-\sqrt{3} \]

1. Start with the equation: \[ 2\cos x = -\sqrt{3} \] 2. Divide both sides by 2 to isolate \(\cos x\): \[ \cos x = -\frac{\sqrt{3}}{2} \] 3. Identify the angles in the interval \( 0 \leq x < 2\pi \) where \(\cos x = -\frac{\sqrt{3}}{2}\). Recall that: - \(\cos x = \frac{\sqrt{3}}{2}\) at \(x = \frac{\pi}{6}\) and \(x = \frac{11\pi}{6}\), - And the cosine function is negative in the second and third quadrants. 4. The corresponding angles in the second and third quadrants are: - Second quadrant: \(x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\) - Third quadrant: \(x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}\) 5. Therefore, the solutions for \( x \) are: \[ x = \frac{5\pi}{6} \quad \text{and} \quad x = \frac{7\pi}{6} \]