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📅 May 14, 2026 👤 scraface

May 14, 2026

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📝 Recent Posts

e 0$ â OK.
$\sin(90^\circ) = 1$, $\cos(90^\circ) = 0$ â $0
e 1$ â Wait! But $\cos z = 0$, so only valid if $2\sin z - 1 = 0$, b...
e 0$. Contradiction?
Wait â at $z = 90^\circ$, $\cos z = 0$, so LHS: $\sin(180^\circ) = 0...
$\sin(2 \cdot 90^\circ) = \sin(180^\circ) = 0$, $\cos(90^\circ) = 0$ â...
$\sin(270^\circ) = -1$, $\cos(270^\circ) = 0$ â $-1 = 0$? No â Not...
So only solutions where $\cos z = 0$ and $\sin z = 1/2$, but $\cos z =...
e 1/2$.
Only valid are $z = 30^\circ$ and $z = 150^\circ$.

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