We can solve these equations with a simple substitution:

For both problems, let x = cos θ. These two problems now become

1) 4x^{2} = 1

2) x^{2} + x - 1 = 0

Now we are back in algebra territory. The first equation 4x^{2} = 1 has solution x = ± ^{1}/_{2}. (I trust you can fill in the details.) Since x = cos θ, we have cos θ = ± ^{1}/_{2 }. For 0°≤θ≤360°, we have four values of θ on the unit circle with cos θ = ± ^{1}/_{2}. The solution set is therefore {60°, 120°, 240°, 300°}.

For the second problem, let's solve x^{2} + x - 1 = 0 by completing the square:

x^{2} + x - 1 = 0 <----> x^{2} + x = 1

<----> x^{2} + x + (^{1}/_{2})^{2 }= 1 + (^{1}/_{2})^{2}

<----> (x + ^{1}/_{2})^{2} = ^{5}/_{4}

<----> x = ^{(-1} ^{± √5)}/_{2}

Remember our substitution x = cos θ ? This means cos θ = ^{(-1} ^{± √5)}/_{2}. Unfortunately, this doesn't match any coordinate listed on the unit circle. So, our final answer is the solution set {cos^{-1} ^{(-1 + √5)}/_{2}, cos^{-1} ^{(-1 - √5)}/_{2}}. For a numerical solution, you must plug these values into a calculator.