\( \left. \begin{array} { l | l | l } { 5.5 } & { \frac { 3 \cos x } { 1 + \sin x } + 3 \tan x = \frac { 3 } { \cos x } } \end{array} \right. \)

We start with the equation \[ \frac{3\cos x}{1+\sin x} + 3\tan x = \frac{3}{\cos x}. \] Since \[ \tan x = \frac{\sin x}{\cos x}, \] we rewrite the equation as \[ \frac{3\cos x}{1+\sin x} + 3\frac{\sin x}{\cos x} = \frac{3}{\cos x}. \] To eliminate the denominators, multiply both sides of the equation by \(\cos x(1+\sin x)\): \[ \cos x(1+\sin x) \cdot \frac{3\cos x}{1+\sin x} + \cos x(1+\sin x) \cdot 3\frac{\sin x}{\cos x} = \cos x(1+\sin x) \cdot \frac{3}{\cos x}. \] Simplify each term: 1. The first term: \[ \frac{3\cos x}{1+\sin x} \cdot \cos x(1+\sin x) = 3\cos^2 x. \] 2. The second term: \[ 3\frac{\sin x}{\cos x} \cdot \cos x(1+\sin x) = 3\sin x(1+\sin x). \] 3. The right-hand side: \[ \frac{3}{\cos x} \cdot \cos x(1+\sin x) = 3(1+\sin x). \] Thus, the equation becomes \[ 3\cos^2 x + 3\sin x(1+\sin x) = 3(1+\sin x). \] Divide both sides by 3: \[ \cos^2 x + \sin x(1+\sin x) = 1+\sin x. \] Expand the term \(\sin x(1+\sin x)\): \[ \cos^2 x + \sin x + \sin^2 x = 1+\sin x. \] Subtract \(\sin x\) from both sides: \[ \cos^2 x + \sin^2 x = 1. \] This is the well-known Pythagorean identity, which holds for all real \(x\). However, recall that we multiplied by \(\cos x(1+\sin x)\). We must ensure that we are not including any values that make these factors zero: 1. \(\cos x \neq 0\) (since division by \(\cos x\) occurs in \(\tan x\) and in the term \(\frac{3}{\cos x}\)). 2. \(1+\sin x \neq 0\) (to avoid division by zero in the term \(\frac{3\cos x}{1+\sin x}\)). This means \(\sin x \neq -1\). Thus, the solution is: All \(x\) such that the identity holds (which it does for all \(x\)) and the restrictions \[ \cos x \neq 0 \quad \text{and} \quad \sin x \neq -1 \] are satisfied. In summary, the solution is: \[ \boxed{\{\,x\in\mathbb{R} : \cos x\neq0 \text{ and } \sin x\neq -1\,\}}. \]