The given function is: \[ f(\theta) = \cos^2 \theta - 6 \sin \theta \cos \theta + 3 \sin^2 \theta + 2 \] By simplifying step by step: 1. Express \( f(\theta) \) as: \[ f(\theta) = 3 + 2 \sin^2 \theta - 3 \sin 2\theta \] where \( \sin 2\theta = 2 \sin \theta \cos \theta \). 2. Rewrite the equation in terms of \( \cos 2\theta \) and \( \sin 2\theta \): \[ f(\theta) = 3 + (1 - \cos 2\theta) - 3 \sin 2\theta \] This simplifies to: \[ f(\theta) = 4 - \cos 2\theta - 3 \sin 2\theta \] Now, to find the maximum and minimum values, consider the expression \[\] \( 4 - \cos 2\theta - 3 \sin 2\theta \) \[\] as a linear combination of cosine and sine terms. \[\] We can express it in the form \( R \cos(2\theta + \alpha) \). \[\] Step 1: Find \( R \) and \( \alpha \) To do this, we write: \[ R \cos(2\theta + \alpha) = R (\cos 2\theta \cos \alpha - \sin 2\theta \sin \alpha) \] Equating this to \( 4 - \cos 2\theta - 3 \sin 2\theta \), we have: \[ R \cos \alpha = -1 \quad \text{and} \quad R \sin \alpha = -3 \] Thus, \( R \) can be found using the Pythagorean identity: \[ R = \sqrt{(-1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} \] Step 2: The maximum value of \( \cos(2\theta + \alpha) \) is 1, and the minimum value is -1. Therefore: \[\] Maximum value of \( 4 - \cos 2\theta - 3 \sin 2\theta \) is: \[ 4 + \sqrt{10} \] Minimum value of \( 4 - \cos 2\theta - 3 \sin 2\theta \) is: \[ 4 - \sqrt{10} \] Thus, the maximum value of the function is \( 4 + \sqrt{10} \), and the minimum value is \( 4 - \sqrt{10} \).