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Trigonometry
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Find all solutions of the equation. & Find the solutions in the interval [0, 2π). csc θ/2 = sin θ/2
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is sec
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\[\csc x = \frac{ 1 }{ \sin x }\] so your equation becomes \[\frac{ 1 }{ \sin \frac{ \theta }{ 2 } } = \sin \frac{ \theta }{ 2 }\] Multiplying gives\[\sin ^{2}\frac{ \theta }{ 2 } = 1\] This has two possibilities:\[\sin \frac{ \theta }{ 2 } = 1 \]or\[\sin \frac{ \theta }{ 2 } = -1\] The solutions are\[\frac{ \theta }{ 2 } = \frac{ 1 }{ 2 }\pi +2k \pi\] and\[\frac{ \theta }{ 2 } = \frac{ 3 }{ 2 }\pi +2k \pi\]We need θ, so we must multiply these solutions with 2: \[\theta= \pi + 4kpi\]\[\theta = 3\pi + 4kpi\]Restricting solutions to the interval [0, 2π), we have\[\theta = \pi\] Done! ZeHanz
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