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

["All Four Solutions Satisfy This Condition? Understood: Algebra, Identity, and the Full Picture", "When solving trigonometric equations involving complex angles, a common question arises:\nDoes the equation $\cos z (2\sin z - 1) = 0$ truly yield all four satisfying cases—$\cos z = 0$ or $\sin z = \frac{1}{2}$—and can all four satisfy simultaneously?", "Let’s carefully explore the mathematics behind this equation, clarify its solutions, and confirm whether all four possibilities—four angles in $[0, 2\pi)$—truly coexist.", "---", "### What Does the Equation Say?", "Start with the identity:", "$$\n\cos z (2\sin z - 1) = 0\n$$", "This expression equals zero when either factor is zero, thanks to the zero product property. So we break it down:", "1. $\cos z = 0$\n2. $2\sin z - 1 = 0$ ⟹ $\sin z = \frac{1}{2}$", "So far, so good: those are the two distinct conditions.", "---", "### The Two Families of Solutions", "#### Case 1: $\cos z = 0$", "The cosine function equals zero at:", "$$\nz = \frac{\pi}{2}, \quad \frac{3\pi}{2} \quad \ ext{within } [0, 2\pi)\n$$", "These are standard solutions from the unit circle.", "#### Case 2: $\sin z = \frac{1}{2}$", "The sine function equals $1/2$ at:", "- In the first quadrant: $z = \frac{\pi}{6}$\n- In the second quadrant: $z = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$", "Thus, $\sin z = \frac{1}{2}$ gives solutions:", "$$\nz = \frac{\pi}{6}, \quad \frac{5\pi}{6}\n$$", "---", "### Are There Exactly Four Solutions?", "At this point, we have:", "- $\cos z = 0$ → $z = \frac{\pi}{2}, \frac{3\pi}{2}$ — 2 solutions\n- $\sin z = \frac{1}{2}$ → $z = \frac{\pi}{6}, \frac{5\pi}{6}$ — 2 solutions", "Total: 4 distinct solutions in $[0, 2\pi)$", "These values are all distinct angles:", "$$\nz = \frac{\pi}{6}, \quad \frac{\pi}{2}, \quad \frac{5\pi}{6}, \quad \frac{3\pi}{2}\n$$", "---", "### Does “All Four Satisfy” Mean They Satisfy This Equation?", "Yes — each of these four angles satisfies:", "$$\n\cos z = 0 \quad \ ext{or} \quad \sin z = \frac{1}{2}\n$$", "So algebraically, all satisfy the equation, and each one individually satisfies the condition expressed by the product being zero.", "Note: The equation does not require all four solutions to be true at once in the sense of being identity components—rather, each solution arises from satisfying one of the two case conditions.", "---", "### Why Not More or Fewer?", "Could there be more solutions?", "- No. The cosine function has exactly two zeros in $[0,2\pi)$, and sine equals $1/2$ exactly twice in that interval.\n- No overlapping or additional roots in this interval.", "Could four solutions satisfy four independent conditions simultaneously? Not here—each solution satisfies a single route (either cosine or sine zero/value), not more.", "---", "### Summary Table: Solutions to $\cos z (2\sin z - 1) = 0$ in $[0, 2\pi)$", "| Condition | Solutions in $[0, 2\pi)$ | Number of Solutions |\n|----------------|---------------------------|---------------------|\n| $\cos z = 0$ | $\frac{\pi}{2}, \frac{3\pi}{2}$ | 2 |\n| $\sin z = \frac{1}{2}$ | $\frac{\pi}{6}, \frac{5\pi}{6}$ | 2 |\n| Total | — | 4 |", "---", "### Final Thoughts on the Algebraic Identity", "This equation exemplifies how trigonometric products zero out via disjunctive logical conditions, not exclusive OR. All four angles are valid because each independently satisfies either factor. Algebra confirms no contradiction—rather, it produces a union of solution sets:", "$$\nz \in \left{ \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2} \right}\n$$", "So yes, all four values satisfy the equation, and collectively they represent all distinct solutions over one full period.", "---", "Key Takeaway:\nWhen dealing with equations of the form $f(z)(g(z)) = 0$, all solutions stem from satisfying at least one factor. Here, $\cos z = 0$ or $\sin z = 1/2$ gives exactly four distinct solutions—showing a clean, consistent application of trigonometric algebra.", "---", "Keywords: trigonometric equation solutions, $\cos z (2\sin z - 1) = 0$, $\cos z = 0 or $\sin z = 1/2$, all four satisfy condition, algebra explanation, sine and cosine identities, periodic functions, complex solutions, trig identity, zero product property.", "---", "For anyone diving into trigonometric equations with complex arguments, remember: break the product into cases, solve each independently, and verify distinctness—then count confidently. All four do satisfy the equation, and each does so via a clean algebraic principle."]