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📅 May 4, 2026 👤 scraface

May 04, 2026

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At $z = 270^\circ$, $\cos z = 0$, so included. But $\sin z = \sin(270^...
e 1/2$, but the equation $\cos z (2\sin z - 1) = 0$ only requires that...
\sin(2z) = \sin(540^\circ) = 0, \quad \cos(z) = \cos(270^\circ) = 0 \R...
So even though $\sin z
e 1/2$, the $\cos z = 0$ makes the whole expression zero. But our fact...
$\cos z = 0$ â $z = 90^\circ, 270^\circ$.
$\sin z = 1/2$ â $z = 30^\circ, 150^\circ$.
All four satisfy the equation.
Thus, the solutions are $30^\circ, 90^\circ, 150^\circ, 270^\circ$.
Question: Compute $(\cos 15^\circ + i \sin 15^\circ)^6$ and express th...

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