Solve \( 3 \operatorname{cosec} A=7 \) to the nearest degree, \( 0 \leq x \leq 360 \)

To solve the equation \( 3 \operatorname{cosec} A = 7 \), we first rewrite it in terms of sine: \[ \operatorname{cosec} A = \frac{1}{\sin A} \] This gives us: \[ 3 \cdot \frac{1}{\sin A} = 7 \] Now, we can solve for \( \sin A \): \[ \frac{3}{\sin A} = 7 \] Multiplying both sides by \( \sin A \) and then by \( \frac{1}{7} \), we have: \[ 3 = 7 \sin A \quad \Rightarrow \quad \sin A = \frac{3}{7} \] Next, we need to find the angle \( A \) whose sine is \( \frac{3}{7} \). We can use the inverse sine function: \[ A = \sin^{-1}\left(\frac{3}{7}\right) \] Calculating \( \sin^{-1}\left(\frac{3}{7}\right) \): \[ A \approx 0.436 \text{ radians} \approx 25.1^\circ \] However, since we need to find angles \( A \) in the range \( 0 \leq A < 360^\circ \), and sine is positive in both the first and second quadrants, we also have: \[ A = 180^\circ - 25.1^\circ \approx 154.9^\circ \] Thus, the solutions for \( A \) in degrees are: \[ A \approx 25^\circ \quad \text{and} \quad A \approx 155^\circ \] So the final solutions are: \[ A \approx 25^\circ \quad \text{and} \quad A \approx 155^\circ \]