\sin(2z) = \sin(540^\circ) = 0, \quad \cos(z) = \cos(270^\circ) = 0 \Rightarrow 0 = 0 ext{ true}.
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- 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:
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- e 1/2$, the $\cos z = 0$ makes the whole expression zero. But our factorization is correct: $\cos z (2\sin z - 1) = 0$ is equivalent to the equation. So solutions when $\cos z = 0$ or $\sin z = 1/2$.
- $\cos z = 0$ â $z = 90^\circ, 270^\circ$.
