a) \( \cos 2 \theta+\cos ^{2} \theta-5 \cos \theta-1=0 \)

Let's redefine \( x = \cos \theta \). The equation becomes: \[ \cos 2\theta + \cos^2 \theta - 5 \cos \theta - 1 = 0. \] Using the double angle formula, \( \cos 2\theta = 2 \cos^2 \theta - 1 \), we can substitute this into the equation: \[ 2x^2 - 1 + x^2 - 5x - 1 = 0. \] Combining like terms yields: \[ 3x^2 - 5x - 2 = 0. \] Now, let's factor or use the quadratic formula to find \( x \): \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{5 \pm \sqrt{(-5)^2 - 4(3)(-2)}}{2(3)}. \] Calculating the discriminant: \[ 25 + 24 = 49, \] \[ \sqrt{49} = 7. \] Thus, we have: \[ x = \frac{5 \pm 7}{6}. \] This gives us two possible solutions for \( x \): 1. \( x = \frac{12}{6} = 2 \) (not valid since \( \cos \theta \) must be between -1 and 1), 2. \( x = \frac{-2}{6} = -\frac{1}{3} \). Now that we have \( \cos \theta = -\frac{1}{3} \). To find \( \theta \), we can use the cosine inverse: \[ \theta = \cos^{-1}\left(-\frac{1}{3}\right). \] This provides possible angles in the ranges of quadrants II and III, specifically: \[ \theta \approx 109.47^\circ \quad \text{and} \quad \theta \approx 250.53^\circ. \] These are the solutions for the original equation!