\( \frac { \cos 2 A } { 3 - \sin 2 A - 4 \sin ^ { 2 } A } = \frac { \tan A + 1 } { \tan A + 3 } \)

To solve the equation \[ \frac { \cos 2 A } { 3 - \sin 2 A - 4 \sin ^ { 2 } A } = \frac { \tan A + 1 } { \tan A + 3 }, \] let's use some trigonometric identities. We know that \[ \cos 2A = 1 - 2\sin^2 A, \] and \[ \sin 2A = 2\sin A \cos A. \] We can also rewrite \(\tan A\) in terms of \(\sin A\) and \(\cos A\): \[ \tan A = \frac{\sin A}{\cos A}. \] Now, use the identities to rewrite \(3 - \sin 2A - 4\sin^2 A\): Expanding \(\sin 2A\), \[ 3 - 2\sin A \cos A - 4\sin^2 A. \] Now, substituting cosine with \(\sqrt{1 - \sin^2 A}\), we need to simplify and combine terms. Additionally, notice that we can analyze both sides of the equation based on the tangent representation. Multiplying both sides by \((\tan A + 3)(3 - \sin 2A - 4\sin^2 A)\) can help eliminate the fraction. From this state, express everything in terms of \(\sin A\) and \(\cos A\) or \(\tan A\) directly. Continue solving for values of \(A\) such that both sides will hold true, eventually reverting back to the unit circle or special angle values as needed. To find specific angles, consider special angles where the sine and cosine values are known, such as \(0\), \(\frac{\pi}{4}\), \(\frac{\pi}{2}\), or even outside the range with periodic properties: - If \(\tan A = 0\) gives \(A = 0\). - Use rational tangent values to compare left to right. From here, the goal is systematic simplification and substitution, focusing on known values or angles until simplifying yields valid \(A\) values. Ultimately, you'll find specific solutions to check back against original equations, maintaining the relationship integrity set by trigonometric functions.