b) \( \cos ^{2} \theta+2 \sin 2 \theta+3 \sin ^{2} \theta=0 \)

To solve the equation \( \cos^{2} \theta + 2 \sin 2 \theta + 3 \sin^{2} \theta = 0 \), we can utilize the double angle identity for sine, which states that \( \sin 2\theta = 2 \sin \theta \cos \theta \). Using this, we can rewrite the equation as: \[ \cos^{2} \theta + 4 \sin \theta \cos \theta + 3 \sin^{2} \theta = 0. \] Next, we can use the Pythagorean identity, \( \sin^{2} \theta = 1 - \cos^{2} \theta \), to express everything in terms of \( \cos^{2} \theta \): \[ \cos^{2} \theta + 4 \sin \theta \cos \theta + 3(1 - \cos^{2} \theta) = 0. \] This becomes: \[ \cos^{2} \theta + 4 \sin \theta \cos \theta + 3 - 3 \cos^{2} \theta = 0, \] which simplifies to: \[ -2 \cos^{2} \theta + 4 \sin \theta \cos \theta + 3 = 0. \] Now, rearranging gives us: \[ 2 \cos^{2} \theta - 4 \sin \theta \cos \theta - 3 = 0. \] This is a quadratic equation in terms of \( \cos \theta \). Let’s denote \( x = \cos \theta \), thus rewriting it yields: \[ 2x^{2} - 4 \sin \theta x - 3 = 0. \] To proceed, we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 2 \), \( b = -4 \sin \theta \), and \( c = -3 \). Now, find the solutions for \( x \) accordingly. For the sake of clarity, let's summarize: \[ x = \frac{4 \sin \theta \pm \sqrt{(4 \sin \theta)^{2} + 24}}{4} = \sin \theta \pm \sqrt{\sin^{2} \theta + 6}. \] You can now find the values of \( \cos \theta \) from the above equations and eventually calculate the angles \( \theta \) that satisfy this trigonometric equation. Happy calculating!