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📅 May 12, 2026 👤 scraface

May 12, 2026

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📝 Recent Posts

$\cos z = 0 \Rightarrow z = 90^\circ, 270^\circ$.
$2\sin z - 1 = 0 \Rightarrow \sin z = rac{1}{2} \Rightarrow z = 30^\c...
All solutions in $[0^\circ, 360^\circ]$ are $z = 30^\circ, 90^\circ, 1...
Check each:
$\sin(60^\circ) = \sqrt{3}/2$, $\cos(30^\circ) = \sqrt{3}/2$ â equal...
$\sin(270^\circ) = -1$, $\cos(270^\circ) = 0$ â $-1
e 0$ 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?

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