Solve the following equation for values of \( \theta \) between \( 0^{\circ} \) and \( 360^{\circ} \) \[ 2-3 \cos ^{2} \theta=2 \sin \theta \]

First, let's rewrite the equation: \[ 2 - 3 \cos^2 \theta = 2 \sin \theta. \] Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can express \( \cos^2 \theta \) as \( 1 - \sin^2 \theta \). Substituting this into the equation gives: \[ 2 - 3(1 - \sin^2 \theta) = 2 \sin \theta. \] Now, simplify the equation: \[ 2 - 3 + 3 \sin^2 \theta = 2 \sin \theta, \] which simplifies to \[ 3 \sin^2 \theta - 2 \sin \theta - 1 = 0. \] Next, we'll let \( x = \sin \theta \), turning the quadratic into: \[ 3x^2 - 2x - 1 = 0. \] Using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): Here \( a = 3, b = -2, c = -1 \): \[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3}, \] \[ x = \frac{2 \pm \sqrt{4 + 12}}{6}, \] \[ x = \frac{2 \pm \sqrt{16}}{6}, \] \[ x = \frac{2 \pm 4}{6}. \] This gives: 1. \( x = \frac{6}{6} = 1 \) 2. \( x = \frac{-2}{6} = -\frac{1}{3} \) Now we can find \( \theta \): 1. For \( x = \sin \theta = 1 \): \[ \theta = 90^\circ. \] 2. For \( x = \sin \theta = -\frac{1}{3} \): To find \( \theta \), we use the arcsine function: \[ \theta = \arcsin\left(-\frac{1}{3}\right). \] This will give us an angle in the fourth quadrant: \[ \theta \approx -19.1^\circ, \] but since we want angles between \( 0^\circ \) and \( 360^\circ \), we add \( 360^\circ \) to get: \[ \theta \approx 360^\circ - 19.1^\circ \approx 340.9^\circ. \] Additionally, there is another solution for sine in the third quadrant: \[ \theta = 180^\circ + 19.1^\circ \approx 199.1^\circ. \] In conclusion, the values of \( \theta \) between \( 0^\circ \) and \( 360^\circ \) that satisfy the equation are: \[ \theta \approx 90^\circ, 199.1^\circ, 340.9^\circ. \]