Question: Find all angles $z \in [0^\circ, 360^\circ]$ that satisfy $\sin(2z) = \cos(z)$.

["Finding All Angles $z \in [0^\circ, 360^\circ]$ That Satisfy $\sin(2z) = \cos(z)$", "Solving trigonometric equations like $\sin(2z) = \cos(z)$ often challenges students, but breaking the problem down step by step makes it manageable. In this article, we’ll explore how to find all angles $z$ within the interval $[0^\circ, 360^\circ]$ that satisfy $\sin(2z) = \cos(z)$.", "---", "### Step 1: Use Trigonometric Identities", "Start by recalling the double-angle identity for sine:", "[\n\sin(2z) = 2\sin(z)\cos(z)\n]", "Substitute this into the original equation:", "[\n2\sin(z)\cos(z) = \cos(z)\n]", "Now rearrange the equation:", "[\n2\sin(z)\cos(z) - \cos(z) = 0\n]", "Factor out $\cos(z)$:", "[\n\cos(z)(2\sin(z) - 1) = 0\n]", "---", "### Step 2: Solve the Factored Equation", "Set each factor equal to zero:", "1. $\cos(z) = 0$", "2. $2\sin(z) - 1 = 0 \Rightarrow \sin(z) = \frac{1}{2}$", "We solve each equation separately within $[0^\circ, 360^\circ]$.", "---", "### Case 1: $\cos(z) = 0$", "The cosine function equals zero at:", "[\nz = 90^\circ \quad \ ext{and} \quad z = 270^\circ\n]", "These are the standard angles in the second and third quadrants where cosine is zero.", "---", "### Case 2: $\sin(z) = \frac{1}{2}$", "We find angles $z$ where sine equals $\frac{1}{2}$ in the interval $[0^\circ, 360^\circ]$. Recall that $\sin(z) = \frac{1}{2}$ at:", "[\nz = 30^\circ \quad \ ext{and} \quad z = 150^\circ\n]", "These lie in the first and second quadrants, where sine is positive.", "---", "### Step 3: Combine All Solutions", "Collect all solutions from both cases:", "[\nz = 30^\circ,\ 90^\circ,\ 150^\circ,\ 270^\circ\n]", "These are all the angles in $[0^\circ, 360^\circ]$ satisfying $\sin(2z) = \cos(z)$.", "---", "### Step 4: Verification (Optional but Recommended)", "Let’s verify a few solutions:", "- For $z = 30^\circ$:\n $\sin(60^\circ) = \frac{\sqrt{3}}{2},\ \cos(30^\circ) = \frac{\sqrt{3}}{2}$ → Equal.", "- For $z = 90^\circ$:\n $\sin(180^\circ) = 0,\ \cos(90^\circ) = 0$ → Equal.", "- For $z = 150^\circ$:\n $\sin(300^\circ) = -\frac{1}{2},\ \cos(150^\circ) = -\frac{\sqrt{3}}{2}$ — wait! This isn’t equal numerically.", "Wait! Let's double-check $z = 150^\circ$.", "$\sin(2 \cdot 150^\circ) = \sin(300^\circ) = -\frac{1}{2}$\n$\cos(150^\circ) = -\frac{\sqrt{3}}{2} \approx -0.866$ — not equal!", "This reveals a mistake.", "---", "### Important Correction: Re-analyze Case 2", "We found $\sin(z) = \frac{1}{2}$ at $z = 30^\circ$ and $150^\circ$. But substituting $z = 150^\circ$ does not satisfy $\sin(2z) = \cos(z)$. This suggests a deeper check is needed.", "Wait — actually, let's carefully re-evaluate:", "We solve $\sin(z) = \frac{1}{2}$:", "- $z = 30^\circ$: $\sin(30^\circ) = 0.5$, $\cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.866$ → $\sin(2z) = \sin(60^\circ) = \frac{\sqrt{3}}{2} = \cos(30^\circ)$. So it does satisfy.", "- $z = 150^\circ$: $\sin(150^\circ) = 0.5$, but $\cos(150^\circ) = -\frac{\sqrt{3}}{2} \approx -0.866$ → $\sin(300^\circ) = -\frac{1}{2} <br/>\ne -\frac{\sqrt{3}}{2}$", "Thus, $z = 150^\circ$ is not a solution, despite $\sin(z) = \frac{1}{2}$. Why?", "Because $\sin(2z) = 2\sin(z)\cos(z)$, not just $\sin(z)$. Though $\sin(z) = \frac{1}{2}$ is correct, the full equation $2\sin(z)\cos(z) = \cos(z)$ requires both signs and values to match.", "So only when $\cos(z) <br/>\ne 0$ do we require:", "[\n2\sin(z) - 1 = 0 \Rightarrow \sin(z) = \frac{1}{2}\n]", "but only if $\cos(z) <br/>\ne 0$, and we must confirm the equality.", "Since $\sin(2z) = \sin(300^\circ) = -\frac{1}{2}$, while $\cos(150^\circ) = -\frac{\sqrt{3}}{2} <br/>\ne -\frac{1}{2}$, the equality fails.", "Therefore, only $z = 30^\circ$ from $\sin(z) = \frac{1}{2}$ satisfies the full equation.", "But earlier, $\cos(z) = 0$ gave $z = 90^\circ, 270^\circ$, both valid.", "So only two valid solutions: $z = 30^\circ, 90^\circ$.", "Wait — but we missed something.", "Let’s go back:", "From $\cos(z)(2\sin(z) - 1) = 0$, we have two cases:", "- $\cos(z) = 0 \Rightarrow z = 90^\circ, 270^\circ$", "- $2\sin(z) - 1 = 0 \Rightarrow \sin(z) = \frac{1}{2} \Rightarrow z = 30^\circ, 150^\circ$", "But must verify each in original equation $\sin(2z) = \cos(z)$", "---", "### Full Verification", "| $z$ | $\sin(2z)$ | $\cos(z)$ | Equal? |\n|-----------|-----------------------|------------------|--------|\n| $30^\circ$ | $\sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.866$ | $\cos(30^\circ) = \frac{\sqrt{3}}{2}$ → Yes |\n| $90^\circ$ | $\sin(180^\circ) = 0$ | $\cos(90^\circ) = 0$ | Yes |\n| $150^\circ$| $\sin(300^\circ) = -\frac{1}{2}$ | $\cos(150^\circ) = -\frac{\sqrt{3}}{2} \approx -0.866$ | No |\n| $270^\circ$| $\sin(540^\circ) = \sin(180^\circ) = 0$ | $\cos(270^\circ) = 0$ | Yes |", "Only $30^\circ, 90^\circ, 270^\circ$ satisfy the equation.", "---", "### Final Answer", "The angles $z \in [0^\circ, 360^\circ]$ that satisfy $\sin(2z) = \cos(z)$ are:", "[\n\boxed{30^\circ,\ 90^\circ,\ 270^\circ}\n]", "---", "### Why This Matters (SEO Keywords)", "- $\sin(2z) = \cos(z)$\n- Trigonometric equation solutions\n- Solve $\sin(2z) = \cos(z)$\n- Angles $z$ in $[0^\circ, 360^\circ]$\n- How to solve trigonometric equations\n- Verify solutions $\sin(2z) = \cos(z)$\n- Step-by-step trigonometric identities\n- Multiple angles identity\n- Solve $2\sin z \cos z = \cos z$", "Optimizing for search: Use long-tail keywords like “solve $\sin(2z) = \cos(z)$ in degrees” and include exact solutions within specified domain to help readers find answers efficiently.", "---", "Summary:\nUse identity $\sin(2z) = 2\sin z \cos z$, rewrite to $\cos z (2\sin z - 1) = 0$, solve each case, verify each solution in the original equation. Valid solutions in $[0^\circ, 360^\circ]$ are $30^\circ$, $90^\circ$, and $270^\circ$."]