cos(θ)=((1))/((2))

{ "query": { "display": "$$\\cos\\left(θ\\right)=\\frac{\\left(1\\right)}{\\left(2\\right)}$$", "symbolab_question": "EQUATION#\\cos(θ)=\\frac{(1)}{(2)}" }, "solution": { "level": "PERFORMED", "subject": "Trigonometry", "topic": "Trig Equations", "subTopic": "Trig Equations", "default": "θ=\\frac{π}{3}+2πn,θ=\\frac{5π}{3}+2πn", "degrees": "θ=60^{\\circ }+360^{\\circ }n,θ=300^{\\circ }+360^{\\circ }n", "meta": { "showVerify": true } }, "steps": { "type": "interim", "title": "$$\\cos\\left(θ\\right)=\\frac{\\left(1\\right)}{\\left(2\\right)}{\\quad:\\quad}θ=\\frac{π}{3}+2πn,\\:θ=\\frac{5π}{3}+2πn$$", "input": "\\cos\\left(θ\\right)=\\frac{\\left(1\\right)}{\\left(2\\right)}", "steps": [ { "type": "interim", "title": "General solutions for $$\\cos\\left(θ\\right)=\\frac{1}{2}$$", "result": "θ=\\frac{π}{3}+2πn,\\:θ=\\frac{5π}{3}+2πn", "steps": [ { "type": "step", "primary": "$$\\cos\\left(x\\right)$$ periodicity table with $$2πn$$ cycle:<br/>$$\\begin{array}{|c|c|c|c|}\\hline x&\\cos(x)&x&\\cos(x)\\\\\\hline 0&1&π&-1\\\\\\hline \\frac{π}{6}&\\frac{\\sqrt{3}}{2}&\\frac{7π}{6}&-\\frac{\\sqrt{3}}{2}\\\\\\hline \\frac{π}{4}&\\frac{\\sqrt{2}}{2}&\\frac{5π}{4}&-\\frac{\\sqrt{2}}{2}\\\\\\hline \\frac{π}{3}&\\frac{1}{2}&\\frac{4π}{3}&-\\frac{1}{2}\\\\\\hline \\frac{π}{2}&0&\\frac{3π}{2}&0\\\\\\hline \\frac{2π}{3}&-\\frac{1}{2}&\\frac{5π}{3}&\\frac{1}{2}\\\\\\hline \\frac{3π}{4}&-\\frac{\\sqrt{2}}{2}&\\frac{7π}{4}&\\frac{\\sqrt{2}}{2}\\\\\\hline \\frac{5π}{6}&-\\frac{\\sqrt{3}}{2}&\\frac{11π}{6}&\\frac{\\sqrt{3}}{2}\\\\\\hline \\end{array}$$" }, { "type": "step", "result": "θ=\\frac{π}{3}+2πn,\\:θ=\\frac{5π}{3}+2πn" } ], "meta": { "interimType": "Trig General Solutions cos 1Eq" } } ], "meta": { "solvingClass": "Trig Equations", "practiceLink": "/practice/trigonometry-practice#area=main&subtopic=Trig%20Equations", "practiceTopic": "Trig Equations" } }, "plot_output": { "meta": { "plotInfo": { "variable": "θ", "plotRequest": "\\cos(θ)-\\frac{(1)}{(2)}" }, "showViewLarger": true } }, "meta": { "showVerify": true } }