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Home » I Didn’t Believe These Aliens Are Judge Reviewers—Do You See What We Found? - Londonproperty

I Didn’t Believe These Aliens Are Judge Reviewers—Do You See What We Found? - Londonproperty

📅 March 14, 2026 👤 scraface

Mar 14, 2026

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📌 $z = 150^\circ$: $\sin(300^\circ) = -\sqrt{3}/2$, $\cos(150^\circ) = -\sqrt{3}/2$ â OK.

📌 $z = 90^\circ$: $\sin(180^\circ) = 0$, $\cos(90^\circ) = 0$ â 0 = 0 â OK.

📌 $z = 270^\circ$: $\sin(540^\circ) = \sin(180^\circ) = 0$, $\cos(270^\circ) = 0$ â 0 = 0 â OK.

📌 So all four satisfy? But earlier algebra says $\cos z (2\sin z - 1) = 0$, so $\cos z = 0$ or $\sin z = 1/2$.

📌 At $z = 270^\circ$, $\cos z = 0$, so included. But $\sin z = \sin(270^\circ) = -1

📌 e 1/2$, but the equation $\cos z (2\sin z - 1) = 0$ only requires that either factor is zero. But when $\cos z = 0$, we must still satisfy the original? Letâs substitute:

📌 \sin(2z) = \sin(540^\circ) = 0, \quad \cos(z) = \cos(270^\circ) = 0 \Rightarrow 0 = 0 ext{ true}.

📌 So even though $\sin z

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