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

February 22, 2026

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

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So all three? Wait: $\sin(2z) = \cos z$:
$z = 30^\circ$: $\sin(60^\circ) = \sqrt{3}/2$, $\cos(30^\circ) = \sqrt{3}/2$ â OK.
$z = 150^\circ$: $\sin(300^\circ) = -\sqrt{3}/2$, $\cos(150^\circ) = -\sqrt{3}/2$ 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

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