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Home » 5 Tequila Shots That Will Blow Your Mind—Paur Ageless Ingredients! - SUpost

5 Tequila Shots That Will Blow Your Mind—Paur Ageless Ingredients! - SUpost

📅 July 6, 2026 👤 scraface

Jul 06, 2026

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📌 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

📌 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 $z = 90^\circ, 270^\circ$.

📌 $\sin z = 1/2$ â $z = 30^\circ, 150^\circ$.

📌 All four satisfy the equation.

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