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\[ \ln(4) \approx 1.3863 \] - Portal da AcÃºstica

📅 February 28, 2026 👤 scraface

Feb 28, 2026

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\cos\alpha = \frac{1}{2}, \quad \sin\alpha = \frac{\sqrt{3}}{2} \Right...
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^\ci...
Let $ \theta = 2z + 60^\circ $, so:
2z + 60^\circ = 30^\circ + 360^\circ k \Rightarrow 2z = -30^\circ + 36...
2z + 60^\circ = 150^\circ + 360^\circ k \Rightarrow 2z = 90^\circ + 36...
Now find all $ z \in [0^\circ, 360^\circ] $:
For $ z = -15^\circ + 180^\circ k $:

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