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R \cos\alpha = 1, \quad R \sin\alpha = \sqrt{3}
R^2 = 1^2 + (\sqrt{3})^2 = 4 \Rightarrow R = 2
\cos\alpha = \frac{1}{2}, \quad \sin\alpha = \frac{\sqrt{3}}{2} \Rightarrow \alpha = 60^\circ
So the equation becomes:
\sin(2z + 60^\circ) = 1 \Rightarrow \sin(2z + 60^\circ) = \frac{1}{2}
The general solution for $ \sin \theta = \frac{1}{2} $ is:
\theta = 30^\circ + 360^\circ k \quad \text{or} \quad \theta = 150^\circ + 360^\circ k
Let $ \theta = 2z + 60^\circ $, so:
2z + 60^\circ = 30^\circ + 360^\circ k \Rightarrow 2z = -30^\circ + 360^\circ k \Rightarrow z = -15^\circ + 180^\circ k
2z + 60^\circ = 150^\circ + 360^\circ k \Rightarrow 2z = 90^\circ + 360^\circ k \Rightarrow z = 45^\circ + 180^\circ k