5-6sin^2(x)-cos(x)=0

{ "query": { "display": "$$5-6\\sin^{2}\\left(x\\right)-\\cos\\left(x\\right)=0$$", "symbolab_question": "EQUATION#5-6\\sin^{2}(x)-\\cos(x)=0" }, "solution": { "level": "PERFORMED", "subject": "Trigonometry", "topic": "Trig Equations", "subTopic": "Trig Equations", "default": "x=\\frac{π}{3}+2πn,x=\\frac{5π}{3}+2πn,x=1.91063…+2πn,x=-1.91063…+2πn", "degrees": "x=60^{\\circ }+360^{\\circ }n,x=300^{\\circ }+360^{\\circ }n,x=109.47122…^{\\circ }+360^{\\circ }n,x=-109.47122…^{\\circ }+360^{\\circ }n", "meta": { "showVerify": true } }, "steps": { "type": "interim", "title": "$$5-6\\sin^{2}\\left(x\\right)-\\cos\\left(x\\right)=0{\\quad:\\quad}x=\\frac{π}{3}+2πn,\\:x=\\frac{5π}{3}+2πn,\\:x=1.91063…+2πn,\\:x=-1.91063…+2πn$$", "input": "5-6\\sin^{2}\\left(x\\right)-\\cos\\left(x\\right)=0", "steps": [ { "type": "interim", "title": "Rewrite using trig identities", "input": "5-\\cos\\left(x\\right)-6\\sin^{2}\\left(x\\right)", "result": "-1-\\cos\\left(x\\right)+6\\cos^{2}\\left(x\\right)=0", "steps": [ { "type": "step", "primary": "Use the Pythagorean identity: $$\\cos^{2}\\left(x\\right)+\\sin^{2}\\left(x\\right)=1$$", "secondary": [ "$$\\sin^{2}\\left(x\\right)=1-\\cos^{2}\\left(x\\right)$$" ], "result": "=5-\\cos\\left(x\\right)-6\\left(1-\\cos^{2}\\left(x\\right)\\right)" }, { "type": "interim", "title": "Simplify $$5-\\cos\\left(x\\right)-6\\left(1-\\cos^{2}\\left(x\\right)\\right):{\\quad}6\\cos^{2}\\left(x\\right)-\\cos\\left(x\\right)-1$$", "input": "5-\\cos\\left(x\\right)-6\\left(1-\\cos^{2}\\left(x\\right)\\right)", "result": "=6\\cos^{2}\\left(x\\right)-\\cos\\left(x\\right)-1", "steps": [ { "type": "interim", "title": "Expand $$-6\\left(1-\\cos^{2}\\left(x\\right)\\right):{\\quad}-6+6\\cos^{2}\\left(x\\right)$$", "input": "-6\\left(1-\\cos^{2}\\left(x\\right)\\right)", "result": "=5-\\cos\\left(x\\right)-6+6\\cos^{2}\\left(x\\right)", "steps": [ { "type": "step", "primary": "Apply the distributive law: $$a\\left(b-c\\right)=ab-ac$$", "secondary": [ "$$a=-6,\\:b=1,\\:c=\\cos^{2}\\left(x\\right)$$" ], "result": "=-6\\cdot\\:1-\\left(-6\\right)\\cos^{2}\\left(x\\right)", "meta": { "practiceLink": "/practice/expansion-practice", "practiceTopic": "Expand Rules" } }, { "type": "step", "primary": "Apply minus-plus rules", "secondary": [ "$$-\\left(-a\\right)=a$$" ], "result": "=-6\\cdot\\:1+6\\cos^{2}\\left(x\\right)" }, { "type": "step", "primary": "Multiply the numbers: $$6\\cdot\\:1=6$$", "result": "=-6+6\\cos^{2}\\left(x\\right)" } ], "meta": { "interimType": "Algebraic Manipulation Expand Title 1Eq", "gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7EuniBxumkib7XYTT7L6GbjD8PgXlmKgFspRJo/rFGenMwViaLUXkeD+JukROhWdjpifpellp95nMxvM7ofaDfO5byrQDQVCXUD0vH/fvOdz8bYA0b6V2RSTOZ7Os9NODEdLCySabqZf8IRyxurbO2YLF6drOTclI884E0jj8Wjs=" } }, { "type": "interim", "title": "Simplify $$5-\\cos\\left(x\\right)-6+6\\cos^{2}\\left(x\\right):{\\quad}6\\cos^{2}\\left(x\\right)-\\cos\\left(x\\right)-1$$", "input": "5-\\cos\\left(x\\right)-6+6\\cos^{2}\\left(x\\right)", "result": "=6\\cos^{2}\\left(x\\right)-\\cos\\left(x\\right)-1", "steps": [ { "type": "step", "primary": "Group like terms", "result": "=-\\cos\\left(x\\right)+6\\cos^{2}\\left(x\\right)+5-6" }, { "type": "step", "primary": "Add/Subtract the numbers: $$5-6=-1$$", "result": "=6\\cos^{2}\\left(x\\right)-\\cos\\left(x\\right)-1" } ], "meta": { "solvingClass": "Solver", "interimType": "Algebraic Manipulation Simplify Title 1Eq" } } ], "meta": { "solvingClass": "Solver", "interimType": "Generic Simplify Specific 1Eq", "gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7E83s2AtM/Y9CHV+DHm+to0z1Hr+qNGX6fTa52q6rYYhV00rpv8+ZC6TM10tVCSHs16VqxivDdlpb+sNsPUV7D9CFwXyJb1z2Eq0SYZBdTZye8ZsoiyEY3kb+hvAdyLDV7kAjP76qW66lOUsURwT0nQ/ByTem8gA1WxeNu5Q0LdcY93nUk237bMH956WOGPsDtmRi0E3eC0He/xESpULhVg==" } } ], "meta": { "interimType": "Trig Rewrite Using Trig identities 0Eq", "gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7UauBdOh5HVc6RwQXBvqtakQMoXro1fR/6ohMM6qaxKuk+FPosJui0EvbDbx83/CqPO0XSKqaWlOK6mHDvVM10YIYWgyd6SbL0eRRsTHoy/YnSKF5/4+51qVY0U4KnLmxRi0mWGJPyueWI92hEBhaGWC2dtJpUluVzz3fU7XCZSrWwPs1+Gw97t4MeuaNjSYTRvemj3GBE2iIDcXU+cR6iI+gxrQ1tCXeKlYIJ1n6NLewiNrEngO+NNvZ9sqNu+2V" } }, { "type": "interim", "title": "Solve by substitution", "input": "-1-\\cos\\left(x\\right)+6\\cos^{2}\\left(x\\right)=0", "result": "\\cos\\left(x\\right)=\\frac{1}{2},\\:\\cos\\left(x\\right)=-\\frac{1}{3}", "steps": [ { "type": "step", "primary": "Let: $$\\cos\\left(x\\right)=u$$", "result": "-1-u+6u^{2}=0" }, { "type": "interim", "title": "$$-1-u+6u^{2}=0{\\quad:\\quad}u=\\frac{1}{2},\\:u=-\\frac{1}{3}$$", "input": "-1-u+6u^{2}=0", "steps": [ { "type": "step", "primary": "Write in the standard form $$ax^{2}+bx+c=0$$", "result": "6u^{2}-u-1=0" }, { "type": "interim", "title": "Solve with the quadratic formula", "input": "6u^{2}-u-1=0", "result": "{u}_{1,\\:2}=\\frac{-\\left(-1\\right)\\pm\\:\\sqrt{\\left(-1\\right)^{2}-4\\cdot\\:6\\left(-1\\right)}}{2\\cdot\\:6}", "steps": [ { "type": "definition", "title": "Quadratic Equation Formula:", "text": "For a quadratic equation of the form $$ax^2+bx+c=0$$ the solutions are <br/>$${\\quad}x_{1,\\:2}=\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}$$" }, { "type": "step", "primary": "For $${\\quad}a=6,\\:b=-1,\\:c=-1$$", "result": "{u}_{1,\\:2}=\\frac{-\\left(-1\\right)\\pm\\:\\sqrt{\\left(-1\\right)^{2}-4\\cdot\\:6\\left(-1\\right)}}{2\\cdot\\:6}" } ], "meta": { "interimType": "Solving The Quadratic Equation With Quadratic Formula Definition 0Eq", "gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7iwxg3mckhqK/Blm2itDdGH6Ql+mTvFVF63lb/DK9BWU6Josm4vlHGC6ZfSA29PX1Gll/cxm5RbaCI0rSS/xReWAaV5nF/5eIY9S6nkRYfn2bzMHvnkBOT7dG9Au4szHVIT6wFkKlgzOQB/a7Xrzxg3P8Fop9R9b2i6F8MsLpCYh6beiNvNSaeH8Z8MpmfRZkL7cw/VQzrBOIH7BZxt2ptQ7upzez8Ly2EObkVpSZ9PxNz4bOy87AdO8OBNpVeSyBS5s9QdRqjQablkEqEWJEWSjvnfPVdJLFLCOMX+1kqOOB7hvmTIwM3CBFcx3XBJWJDqci355gpRQ8PTS+C1L3Z+GyYgHNIRx5JCuhuidivkeN7xD4NDVhP2bkEDxEeqtesKWbC8mU55vvv8Kzs4kMCcKlXW6oegp4Qt8CxuTsXZ6YQBPL5n44Ah6WDMt/yIbfip83AruRFVHYBQ7iD8OTBYNmnreEEaGOyS+4N+t9v1n4zjvBTCh95txzZviGUMMJeB3JNIgAt4okc53Rulv5qSxd/3OQYxmsr5R6AryEWpbNtW+xt7YXAHnnUXfOhpt2FylN12FcHo+l2d9qBsS/3H5z9JkYomU9P1YGTFRFqDdN5Aod6Hr1Lp2e/29KhSgUFdl3XxkGszefj4hqCRIN0mp6GT0oXcCVg7/8De2Bm0CJqVxX90jlMfh9fKn6dzC4" } }, { "type": "interim", "title": "$$\\sqrt{\\left(-1\\right)^{2}-4\\cdot\\:6\\left(-1\\right)}=5$$", "input": "\\sqrt{\\left(-1\\right)^{2}-4\\cdot\\:6\\left(-1\\right)}", "result": "{u}_{1,\\:2}=\\frac{-\\left(-1\\right)\\pm\\:5}{2\\cdot\\:6}", "steps": [ { "type": "step", "primary": "Apply rule $$-\\left(-a\\right)=a$$", "result": "=\\sqrt{\\left(-1\\right)^{2}+4\\cdot\\:6\\cdot\\:1}" }, { "type": "interim", "title": "$$\\left(-1\\right)^{2}=1$$", "input": "\\left(-1\\right)^{2}", "steps": [ { "type": "step", "primary": "Apply exponent rule: $$\\left(-a\\right)^{n}=a^{n},\\:$$if $$n$$ is even", "secondary": [ "$$\\left(-1\\right)^{2}=1^{2}$$" ], "result": "=1^{2}" }, { "type": "step", "primary": "Apply rule $$1^{a}=1$$", "result": "=1" } ], "meta": { "solvingClass": "Solver", "interimType": "Solver", "gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s78E1FVQW6YvXK7raPRxih+c0ag8T1MwTer44+aCS/ZFAdx7pcd1x/bAWpIL8hAintf05A2GsVmPba4FjoW22b4iKyMg44e9p5G7GRfJ2en9g=" } }, { "type": "interim", "title": "$$4\\cdot\\:6\\cdot\\:1=24$$", "input": "4\\cdot\\:6\\cdot\\:1", "steps": [ { "type": "step", "primary": "Multiply the numbers: $$4\\cdot\\:6\\cdot\\:1=24$$", "result": "=24" } ], "meta": { "solvingClass": "Solver", "interimType": "Solver", "gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s75myq+NgS1FY3DAMtFQcd/GiBFxjfQHIgU3vZ8lSvjvOjkVi15I8rBefLi4Iyt2wrA5mNI8d524k0MHHn76yyMdva/pqcNZ0RI8iVB4DAvzGskbG7YG1bWnmUDxTvw1p+" } }, { "type": "step", "result": "=\\sqrt{1+24}" }, { "type": "step", "primary": "Add the numbers: $$1+24=25$$", "result": "=\\sqrt{25}" }, { "type": "step", "primary": "Factor the number: $$25=5^{2}$$", "result": "=\\sqrt{5^{2}}" }, { "type": "step", "primary": "Apply radical rule: $$\\sqrt[n]{a^n}=a$$", "secondary": [ "$$\\sqrt{5^{2}}=5$$" ], "result": "=5", "meta": { "practiceLink": "/practice/radicals-practice", "practiceTopic": "Radical Rules" } } ], "meta": { "solvingClass": "Solver", "interimType": "Solver", "gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7gcCSkwHGkSzBJS28+axO633X8i0eFrzLJeQiok3gpGoAlilG71elit3w1IBbYN0PPMIzKJkow6rtuXltoCsjrqN6Hv6MoTMtvtU0IQwXdn8tVZVn3juMEEyoNr8u6sPO7uWG86Z/CN1PiruWiWSTwiS3daIZHtloJpe/PvtsyNI=" } }, { "type": "step", "primary": "Separate the solutions", "result": "{u}_{1}=\\frac{-\\left(-1\\right)+5}{2\\cdot\\:6},\\:{u}_{2}=\\frac{-\\left(-1\\right)-5}{2\\cdot\\:6}" }, { "type": "interim", "title": "$$u=\\frac{-\\left(-1\\right)+5}{2\\cdot\\:6}:{\\quad}\\frac{1}{2}$$", "input": "\\frac{-\\left(-1\\right)+5}{2\\cdot\\:6}", "steps": [ { "type": "step", "primary": "Apply rule $$-\\left(-a\\right)=a$$", "result": "=\\frac{1+5}{2\\cdot\\:6}" }, { "type": "step", "primary": "Add the numbers: $$1+5=6$$", "result": "=\\frac{6}{2\\cdot\\:6}" }, { "type": "step", "primary": "Multiply the numbers: $$2\\cdot\\:6=12$$", "result": "=\\frac{6}{12}" }, { "type": "step", "primary": "Cancel the common factor: $$6$$", "result": "=\\frac{1}{2}" } ], "meta": { "solvingClass": "Solver", "interimType": "Solver", "gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7rYEb5H2UhMed/LULBMRUu3b3pfjq4F6GeQrpz1mNPWwgJ/ZZA32ZInFBpDtxBfiKXYGCmiBF99lesmXZ9iIfJ6AZ5He/JUOeeEQ1/UOV2/+iuu1BULe4p4TCP1Dge8oR0VYeMMQvgH2X0uHbf1Jm7YmpXFf3SOUx+H18qfp3MLg=" } }, { "type": "interim", "title": "$$u=\\frac{-\\left(-1\\right)-5}{2\\cdot\\:6}:{\\quad}-\\frac{1}{3}$$", "input": "\\frac{-\\left(-1\\right)-5}{2\\cdot\\:6}", "steps": [ { "type": "step", "primary": "Apply rule $$-\\left(-a\\right)=a$$", "result": "=\\frac{1-5}{2\\cdot\\:6}" }, { "type": "step", "primary": "Subtract the numbers: $$1-5=-4$$", "result": "=\\frac{-4}{2\\cdot\\:6}" }, { "type": "step", "primary": "Multiply the numbers: $$2\\cdot\\:6=12$$", "result": "=\\frac{-4}{12}" }, { "type": "step", "primary": "Apply the fraction rule: $$\\frac{-a}{b}=-\\frac{a}{b}$$", "result": "=-\\frac{4}{12}" }, { "type": "step", "primary": "Cancel the common factor: $$4$$", "result": "=-\\frac{1}{3}" } ], "meta": { "solvingClass": "Solver", "interimType": "Solver", "gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7xrubhzl7D6ENB059zrTkiHb3pfjq4F6GeQrpz1mNPWwgJ/ZZA32ZInFBpDtxBfiKmWiTEpQjat3SO7/l2m58l81j2ZZIW8Tm92w7y1ZxW5PFyoUUXuvn7kSsC/X/60sUsbDR34UYeA7bleA8Apv0UnUv0jwU90/Zz1GhYSW3uIE=" } }, { "type": "step", "primary": "The solutions to the quadratic equation are:", "result": "u=\\frac{1}{2},\\:u=-\\frac{1}{3}" } ], "meta": { "solvingClass": "Equations", "interimType": "Equations" } }, { "type": "step", "primary": "Substitute back $$u=\\cos\\left(x\\right)$$", "result": "\\cos\\left(x\\right)=\\frac{1}{2},\\:\\cos\\left(x\\right)=-\\frac{1}{3}" } ], "meta": { "interimType": "Substitution Method 0Eq" } }, { "type": "interim", "title": "$$\\cos\\left(x\\right)=\\frac{1}{2}{\\quad:\\quad}x=\\frac{π}{3}+2πn,\\:x=\\frac{5π}{3}+2πn$$", "input": "\\cos\\left(x\\right)=\\frac{1}{2}", "steps": [ { "type": "interim", "title": "General solutions for $$\\cos\\left(x\\right)=\\frac{1}{2}$$", "result": "x=\\frac{π}{3}+2πn,\\:x=\\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": "x=\\frac{π}{3}+2πn,\\:x=\\frac{5π}{3}+2πn" } ], "meta": { "interimType": "Trig General Solutions cos 1Eq" } } ], "meta": { "interimType": "N/A" } }, { "type": "interim", "title": "$$\\cos\\left(x\\right)=-\\frac{1}{3}{\\quad:\\quad}x=\\arccos\\left(-\\frac{1}{3}\\right)+2πn,\\:x=-\\arccos\\left(-\\frac{1}{3}\\right)+2πn$$", "input": "\\cos\\left(x\\right)=-\\frac{1}{3}", "steps": [ { "type": "interim", "title": "Apply trig inverse properties", "input": "\\cos\\left(x\\right)=-\\frac{1}{3}", "result": "x=\\arccos\\left(-\\frac{1}{3}\\right)+2πn,\\:x=-\\arccos\\left(-\\frac{1}{3}\\right)+2πn", "steps": [ { "type": "step", "primary": "General solutions for $$\\cos\\left(x\\right)=-\\frac{1}{3}$$", "secondary": [ "$$\\cos\\left(x\\right)=-a\\quad\\Rightarrow\\quad\\:x=\\arccos\\left(-a\\right)+2πn,\\:\\quad\\:x=-\\arccos\\left(-a\\right)+2πn$$" ], "result": "x=\\arccos\\left(-\\frac{1}{3}\\right)+2πn,\\:x=-\\arccos\\left(-\\frac{1}{3}\\right)+2πn" } ], "meta": { "interimType": "Trig Apply Inverse Props 0Eq" } } ], "meta": { "interimType": "N/A" } }, { "type": "step", "primary": "Combine all the solutions", "result": "x=\\frac{π}{3}+2πn,\\:x=\\frac{5π}{3}+2πn,\\:x=\\arccos\\left(-\\frac{1}{3}\\right)+2πn,\\:x=-\\arccos\\left(-\\frac{1}{3}\\right)+2πn" }, { "type": "step", "primary": "Show solutions in decimal form", "result": "x=\\frac{π}{3}+2πn,\\:x=\\frac{5π}{3}+2πn,\\:x=1.91063…+2πn,\\:x=-1.91063…+2πn" } ], "meta": { "solvingClass": "Trig Equations", "practiceLink": "/practice/trigonometry-practice#area=main&subtopic=Trig%20Equations", "practiceTopic": "Trig Equations" } }, "plot_output": { "meta": { "plotInfo": { "variable": "x", "plotRequest": "5-6\\sin^{2}(x)-\\cos(x)" }, "showViewLarger": true } }, "meta": { "showVerify": true } }