b) \( 2 \tan ^{2} \alpha+3 \sec \alpha=0 \)

To solve the equation \( 2 \tan^2 \alpha + 3 \sec \alpha = 0 \), we can first rewrite the tangent and secant functions in terms of sine and cosine: \[ \tan^2 \alpha = \frac{\sin^2 \alpha}{\cos^2 \alpha} \quad \text{and} \quad \sec \alpha = \frac{1}{\cos \alpha}. \] So, the equation becomes: \[ 2 \frac{\sin^2 \alpha}{\cos^2 \alpha} + \frac{3}{\cos \alpha} = 0. \] Multiplying through by \( \cos^2 \alpha \) (noting that \( \cos \alpha \neq 0 \)) gives: \[ 2 \sin^2 \alpha + 3 \cos \alpha = 0. \] Using the identity \( \sin^2 \alpha = 1 - \cos^2 \alpha \): \[ 2(1 - \cos^2 \alpha) + 3 \cos \alpha = 0. \] This simplifies to: \[ 2 - 2 \cos^2 \alpha + 3 \cos \alpha = 0, \] or rearranging gives us a quadratic equation in terms of \( \cos \alpha \): \[ 2 \cos^2 \alpha - 3 \cos \alpha - 2 = 0. \] We can apply the quadratic formula \( \cos \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 2, b = -3, c = -2 \): \[ \cos \alpha = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2} = \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm 5}{4}. \] This gives us two potential solutions: \[ \cos \alpha = \frac{8}{4} = 2 \quad \text{(not possible since } |\cos \alpha| \leq 1\text{)}, \] \[ \cos \alpha = \frac{-2}{4} = -\frac{1}{2}. \] Now, \( \cos \alpha = -\frac{1}{2} \) corresponds to angles in the second and third quadrants: \[ \alpha = 120^\circ + 360^\circ n \quad \text{or} \quad \alpha = 240^\circ + 360^\circ n \quad (n \in \mathbb{Z}). \] Thus, the angles that satisfy the given equation are \( 120^\circ + 360^\circ n \) and \( 240^\circ + 360^\circ n \).