We can simplify the equation by converting everything to sines and cosines: $\frac{\sin x}{\cos x} + \frac{1}{\cos x} = 2\cos x$ Multiplying through by $\cos x$ gives: $\sin x + 1 = 2\cos^2 x$ Using the identity $\cos^2 x + \sin^2 x = 1$, we can substitute $\sin^2 x$ with $1 - \cos^2 x$ to get: $\sin x + 1 = 2 - 2\sin^2 x$ Rearranging terms gives: $2\sin^2 x + \sin x - 1 = 0$ This is a quadratic equation in $\sin x$, which we can solve using the quadratic formula: $\sin x = \frac{-1 \pm \sqrt{1 + 8}}{4}$ The discriminant is positive, so there are two solutions for $\sin x$: $\sin x = \frac{-1 \pm \sqrt{9}}{4} = -1, \frac{1}{2}$ For $\sin x = -1$, we have $x = \frac{3\pi}{2}$. For $\sin x = \frac{1}{2}$, we have $x = \frac{\pi}{6}, \frac{5\pi}{6}$. Therefore, there are a total of $\boxed{3}$ solutions in the interval $\left[ 0, 2\pi \right]$.