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Home » A_{\textremaining} = \pi (4)^2 = 16\pi \text square centimeters. - Esdistancia

A_{\textremaining} = \pi (4)^2 = 16\pi \text square centimeters. - Esdistancia

📅 March 17, 2026 👤 scraface

Mar 17, 2026

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So all three? Wait: $\sin(2z) = \cos z$:
$z = 30^\circ$: $\sin(60^\circ) = \sqrt{3}/2$, $\cos(30^\circ) = \sqrt...
$z = 150^\circ$: $\sin(300^\circ) = -\sqrt{3}/2$, $\cos(150^\circ) = -...
$z = 90^\circ$: $\sin(180^\circ) = 0$, $\cos(90^\circ) = 0$ â 0 = 0...
$z = 270^\circ$: $\sin(540^\circ) = \sin(180^\circ) = 0$, $\cos(270^\c...
So all four satisfy? But earlier algebra says $\cos z (2\sin z - 1) =...
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

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