Solve the equation $3\cos A + 3\sec A + 7\tan A = 0$ for $0^\circ < A < 360^\circ$.

1. Express the equation in terms of sine and cosine

$3\cos A + \frac{3}{\cos A} + \frac{7\sin A}{\cos A} = 0$
Multiply throughout by $\cos A$ (assuming $\cos A \neq 0$):
$3\cos^2 A + 3 + 7\sin A = 0$

2. Convert to a quadratic equation in terms of $\sin A$

Using the identity $\cos^2 A = 1 - \sin^2 A$:
$3(1 - \sin^2 A) + 3 + 7\sin A = 0$
$3 - 3\sin^2 A + 3 + 7\sin A = 0$
$3\sin^2 A - 7\sin A - 6 = 0$

3. Solve the quadratic equation for $\sin A$

$(3\sin A + 2)(\sin A - 3) = 0$
$\sin A = -\frac{2}{3} \quad \text{or} \quad \sin A = 3$
Reject $\sin A = 3$ since $-1 \leqslant \sin A \leqslant 1$.

4. Find the basic angle and determine the quadrants

Basic angle $\alpha = \sin^{-1}\left(\frac{2}{3}\right) \approx 41.810^\circ$
Since $\sin A < 0$, $A$ lies in the 3rd and 4th quadrants.

5. Calculate the final values of $A$

$A = 180^\circ + 41.810^\circ = 221.810^\circ$
$A = 360^\circ - 41.810^\circ = 318.190^\circ$
Rounding to 1 decimal place:
$A = 221.8^\circ, 318.2^\circ$