Find all degree solutions. (Enter your answers as a comma-separated list. Let k be any integer.)2 cos2 4𝜃 + 3 cos 4𝜃 + 1 = 0

This equation is a quadratic in terms of cos(4θ). We can solve it by using the quadratic formula, which is x = [-b ± sqrt(b^2 - 4ac)] / (2a). Here, a = 2, b = 3, and c = 1.

Step 1: Calculate the discriminant (b^2 - 4ac) = (3)^2 - 4*(2)*(1) = 9 - 8 = 1

Step 2: Substitute a, b, and the discriminant into the quadratic formula to find the roots. x = [-3 ± sqrt(1)] / (2*2) = [-3 ± 1] / 4 So, the roots are x = -1 and x = -0.5.

Step 3: Set each root equal to cos(4θ) and solve for θ. If cos(4θ) = -1, then 4θ = arccos(-1) = 180° + 360°k or -180° + 360°k. Dividing by 4 gives θ = 45° + 90°k or -45° + 90°k.

If cos(4θ) = -0.5, then 4θ = arccos(-0.5) = 120° + 360°k or -120° + 360°k. Dividing by 4 gives θ = 30° + 90°k or -30° + 90°k.

So, the solutions to the equation are θ = 45° + 90°k, -45° + 90°k, 30° + 90°k, -30° + 90°k, where k is any integer.