EXERCISE 10 1. Solve for \( \theta \) in the following. Round off your answe (a) \( \tan \theta=2 \), and \( \theta \in\left(0^{\circ} ; 360^{\circ}\right) \) (b) \( \cos \theta=0,657 \), and \( \theta \in\left(0^{\circ} ; 360^{\circ}\right) \) (c) \( \sin \theta=-0,56 \), and \( 0^{\circ}<\theta<360^{\circ} \) (d) \( 3 \cos \theta=-2 \), and \( \theta \in\left(0^{\circ} ; 360^{\circ}\right) \) (e) \( 4 \sin \theta-3=0 \), and \( 0^{\circ}<\theta<360^{\circ} \) (f) \( 2 \tan \theta-0,82=0 \), and \( \theta \in\left(0^{\circ} ; 180^{\circ}\right) \) 2. Solve for \( A \) if \( \sin A=\tan 322^{\circ} \), and \( A \in\left(0^{\circ} ; 360^{\circ}\right) \)

1. (a) We have \[ \tan\theta = 2. \] The principal solution is \[ \theta_0 = \arctan(2) \approx 63.43^\circ. \] Since the period of the tangent function is \(180^\circ\), the general solutions are \[ \theta = 63.43^\circ + 180^\circ k, \quad k\in\mathbb{Z}. \] Within the interval \(\left(0^\circ,\,360^\circ\right)\) the two solutions are: \[ \theta \approx 63.43^\circ \quad\text{and}\quad \theta \approx 63.43^\circ + 180^\circ = 243.43^\circ. \] 1. (b) We have \[ \cos\theta = 0.657. \] The principal solution is \[ \theta_0 = \arccos(0.657) \approx 49.0^\circ. \] Since cosine is positive in quadrants I and IV, the solutions are \[ \theta \approx 49.0^\circ \quad\text{and}\quad \theta \approx 360^\circ - 49.0^\circ = 311.0^\circ. \] 1. (c) We have \[ \sin\theta = -0.56. \] The reference angle is \[ \alpha = \arcsin(0.56) \approx 34^\circ. \] Since sine is negative in quadrants III and IV, the solutions are: \[ \theta \approx 180^\circ + 34^\circ = 214^\circ, \] \[ \theta \approx 360^\circ - 34^\circ = 326^\circ. \] 1. (d) We have \[ 3\cos\theta=-2 \quad\Longrightarrow\quad \cos\theta = -\frac{2}{3} \approx -0.667. \] The reference angle is \[ \alpha = \arccos\left(\frac{2}{3}\right) \approx 48.19^\circ. \] Since cosine is negative in quadrants II and III, the solutions are: \[ \theta \approx 180^\circ - 48.19^\circ = 131.81^\circ, \] \[ \theta \approx 180^\circ + 48.19^\circ = 228.19^\circ. \] 1. (e) We have \[ 4\sin\theta - 3 = 0 \quad\Longrightarrow\quad \sin\theta = \frac{3}{4} = 0.75. \] The principal solution is \[ \theta_0 = \arcsin(0.75) \approx 48.59^\circ. \] Since sine is positive in quadrants I and II, the solutions are: \[ \theta \approx 48.59^\circ \quad\text{and}\quad \theta \approx 180^\circ - 48.59^\circ = 131.41^\circ. \] 1. (f) We have \[ 2\tan\theta - 0.82 = 0 \quad\Longrightarrow\quad \tan\theta = 0.41. \] The principal solution is \[ \theta = \arctan(0.41) \approx 22.3^\circ. \] Since the period of tangent is \(180^\circ\) and the interval is \(\left(0^\circ,\,180^\circ\right)\), the unique solution is: \[ \theta \approx 22.3^\circ. \] 2. We are to solve for \(A\) if \[ \sin A = \tan 322^\circ, \] with \(A \in \left(0^\circ,\,360^\circ\right)\). First, note that \[ 322^\circ = 360^\circ - 38^\circ, \] so \[ \tan 322^\circ = \tan(-38^\circ) = -\tan 38^\circ. \] Calculating, \[ \tan 38^\circ \approx 0.7813, \] thus, \[ \tan 322^\circ \approx -0.7813. \] The equation becomes \[ \sin A = -0.7813. \] The reference angle is \[ \alpha = \arcsin(0.7813) \approx 51.3^\circ. \] Since sine is negative in quadrants III and IV, the solutions for \(A\) are: \[ A \approx 180^\circ + 51.3^\circ = 231.3^\circ, \] \[ A \approx 360^\circ - 51.3^\circ = 308.7^\circ. \]