{"id":79047,"date":"2017-07-16T18:09:41","date_gmt":"2017-07-16T21:09:41","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=79047"},"modified":"2023-04-07T05:44:43","modified_gmt":"2023-04-07T08:44:43","slug":"problemao-equacao-trigonometrica","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problemao-equacao-trigonometrica\/","title":{"rendered":".Problem\u00e3o: Equa\u00e7\u00e3o Trigonom\u00e9trica"},"content":{"rendered":"<p style=\"margin:20px\">\n<div class=\"g1n\" style=\"line-height:25px;\">\n<p class=\"tit\" style=\"line-height:25px;margin:10px\">Problema<br \/>\n<span style=\"font-size:12px;color:#800000\"> (Indicado a partir do 2\u00ba ano do E. M.)<\/span><\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nCalcule, em radianos, a soma [tex]S[\/tex] das ra\u00edzes da equa\u00e7\u00e3o [tex]\\boxed{1 + \\cos(x) + \\cos(2x) + \\cos(3x) = 0}[\/tex], no intervalo [tex][0, \\pi][\/tex].<\/p>\n<p style=\"margin:10px\">\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"g1n\" style=\"line-height:35px;\">\n<p class=\"tit\" style=\"line-height:30px;margin:10px\">Solu\u00e7\u00e3o 1<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nNesta solu\u00e7\u00e3o, usaremos uma das transforma\u00e7\u00f5es de soma em produto da trigonometria: <\/p>\n<ul>\n<li>[tex]\\cos(p) + \\cos(q) = 2 \\cdot \\cos\\left(\\dfrac{p + q}{2}\\right) \\cdot \\cos\\left(\\dfrac{p &#8211; q}{2}\\right)[\/tex].<\/li>\n<\/ul>\n<p>Assim, segue que:<br \/>\n[tex]\\qquad 1 + \\cos(x) + \\cos(2x) + \\cos(3x) = 0[\/tex]<br \/>\n[tex]\\qquad  \\cos(0) + \\cos(x) + \\cos(2x) + \\cos(3x) = 0[\/tex]<br \/>\n[tex]\\qquad  2 \\cdot \\cos\\left(\\dfrac{x + 0}{2}\\right) \\cdot \\cos\\left(\\dfrac{x &#8211; 0}{2}\\right) + 2 \\cdot \\cos\\left(\\dfrac{3x + 2x}{2}\\right) \\cdot \\cos\\left(\\dfrac{3x &#8211; 2x}{2}\\right) = 0[\/tex]<br \/>\n[tex]\\qquad  2 \\cdot \\cos\\left(\\dfrac{x}{2}\\right) \\cdot \\cos\\left(\\dfrac{x}{2}\\right) + 2 \\cdot \\cos\\left(\\dfrac{5x}{2}\\right) \\cdot \\cos\\left(\\dfrac{x}{2}\\right) = 0[\/tex]<br \/>\n[tex]\\qquad 2 \\cdot \\cos\\left(\\dfrac{x}{2}\\right) \\cdot \\left(\\cos\\left(\\dfrac{x}{2}\\right) + \\cos\\left(\\dfrac{5x}{2}\\right) \\right) = 0[\/tex]<br \/>\n[tex]\\qquad 2 \\cdot \\cos\\left(\\dfrac{x}{2}\\right) \\cdot 2 \\cdot \\cos\\left(\\dfrac{\\dfrac{5x}{2} + \\dfrac{x}{2}}{2}\\right) \\cdot \\cos\\left(\\dfrac{\\dfrac{5x}{2} &#8211; \\dfrac{x}{2}}{2}\\right) = 0[\/tex]<br \/>\n[tex]\\qquad 4 \\cdot \\cos\\left(\\dfrac{x}{2}\\right) \\cdot \\cos\\left(\\dfrac{3x}{2}\\right) \\cdot \\cos(x) = 0,[\/tex]<br \/>\ndonde conclu\u00edmos que<br \/>\n[tex]\\qquad  \\boxed{\\cos\\left(\\dfrac{x}{2}\\right) = 0}\\,\\,[\/tex] ou [tex]\\,\\,\\boxed{\\cos\\left(\\dfrac{3x}{2}\\right) = 0}\\,\\,[\/tex] ou [tex]\\,\\,\\boxed{\\cos(x) = 0}[\/tex]<br \/>\nAnalisemos essas tr\u00eas alternativas:<\/p>\n<ul>\n<li>[tex]\\cos\\left(\\dfrac{x}{2}\\right) = 0 \\,\\,\\Leftrightarrow\\,\\, \\dfrac{x}{2} = \\dfrac{\\pi}{2} + k \\pi \\,\\,\\Leftrightarrow\\,\\, x = \\pi + 2k \\pi[\/tex]<\/li>\n<li>[tex]\\cos\\left(\\dfrac{3x}{2}\\right) = 0 \\,\\,\\Leftrightarrow\\,\\, \\dfrac{3x}{2} = \\dfrac{\\pi}{2} + k \\pi \\,\\,\\Leftrightarrow\\,\\, x = \\dfrac{\\pi}{3} + \\dfrac{2k \\pi}{3}[\/tex]<\/li>\n<li>[tex]\\cos(x) = 0 \\,\\,\\Leftrightarrow\\,\\, x = \\dfrac{\\pi}{2} + k \\pi[\/tex].<\/li>\n<\/ul>\n<p>Dessa forma, no intervalo [tex][0 , \\pi][\/tex], obtemos respectivamente as solu\u00e7\u00f5es <\/p>\n<ul>\n<li>[tex]\\boxed{x = \\pi}[\/tex]<\/li>\n<li>[tex]\\boxed{x = \\dfrac{\\pi}{3}}[\/tex]<\/li>\n<li>[tex]\\boxed{x = \\dfrac{\\pi}{2}}[\/tex] <\/li>\n<\/ul>\n<p>e, portanto, [tex]\\fcolorbox{black}{#eee0e5}{$S = \\dfrac{11 \\pi}{6}$}[\/tex].<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 15px;border: 0 none\"\/>\n<p class=\"g2\" style=\"line-height:20px;\">Solu\u00e7\u00e3o elaborada pelos <strong>Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"g1n\" style=\"line-height:30px;\">\n<p class=\"tit\" style=\"line-height:30px;margin:10px\">Solu\u00e7\u00e3o 2<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nTemos a equa\u00e7\u00e3o<br \/>\n[tex]1+cos(x) +cos(3x)+cos(2x)=0 \\quad \\color{#800000}{(i)}[\/tex]<br \/>\ne sabemos que <\/p>\n<ul>\n<li>[tex]cos(2 x) = cos^2 (x) \u2013 sen^2 (x)= cos^2 (x) \u2013 (1-cos^2 (x)) = 2cos^2(x) \u2013 1. \\quad \\color{#800000}{(ii)}[\/tex] <\/li>\n<\/ul>\n<p>Substituindo [tex]\\color{#800000}{(ii)}[\/tex] em [tex]\\color{#800000}{(i)}[\/tex], obtemos:<br \/>\n[tex]1+cos (x) + cos (3x) + 2cos^2 (x) -1 = 0.\\quad \\color{#800000}{(iii)}[\/tex]<br \/>\nSabemos tamb\u00e9m que <\/p>\n<ul>\n<li>[tex]cos(3x) = 4cos^3(x) \u2013 3cos (x). \\quad \\color{#800000}{(iv)}[\/tex] <\/li>\n<\/ul>\n<p>Substituindo [tex]\\color{#800000}{(iv)}[\/tex] em [tex]\\color{#800000}{(iii)}[\/tex], obtemos:<br \/>\n[tex]cos (x)+2cos^2(x) + 4 cos^3 (x) \u2013 3cos (x) =0. \\quad \\color{#800000}{(v)}[\/tex]<br \/>\nFazendo [tex]cos(x)=y[\/tex] em [tex]\\color{#800000}{(v)}[\/tex], resulta a equa\u00e7\u00e3o [tex] \\, 4y^3 + 2y^2 \u2013 2y =0[\/tex], mas<br \/>\n[tex]\\qquad 4y^3 + 2y^2 \u2013 2y =0 \\Leftrightarrow y( 4y^2 + 2y -2) = 0\\Leftrightarrow y( 2y^2 + y -1 ) =0[\/tex],<br \/>\nlogo [tex]y=0[\/tex] ou [tex]2y^2 + y-1 = 0[\/tex].<br \/>\nA equa\u00e7\u00e3o [tex]2y^2 + y-1 = 0[\/tex] nos fornece [tex]y = 1\/2[\/tex] ou [tex]y = -1[\/tex], logo temos as seguintes alternativas:<\/p>\n<ul>\n<li>[tex]cos(x) = 0[\/tex] e [tex]x= \\pi\/2[\/tex]<\/li>\n<li>[tex]cos(x)=-1[\/tex] e [tex]x = \\pi[\/tex]<\/li>\n<li>[tex]cos(x)= 1\/2[\/tex] e [tex]x =\\pi\/3[\/tex].<\/li>\n<\/ul>\n<p>Portanto, [tex]\\boxed{S = \\pi\/2 +\\pi + \\pi\/3= 11\\pi \/ 6}[\/tex]<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 15px;border: 0 none\"\/>\n<p class=\"g2\" style=\"line-height:20px;\">Solu\u00e7\u00e3o elaborada pelo Clube <strong>Descendentes de Pit\u00e1goras<\/strong>, com contribui\u00e7\u00f5es dos <strong>Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n<p style=\"margin:20px\">\n<div class=\"g3\">\nParticipou da discuss\u00e3o os Clubes: <strong>Descendentes de Pit\u00e1goras<\/strong>.<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Problema (Indicado a partir do 2\u00ba ano do E. M.) Calcule, em radianos, a soma [tex]S[\/tex] das ra\u00edzes da equa\u00e7\u00e3o [tex]\\boxed{1 + \\cos(x) + \\cos(2x) + \\cos(3x) = 0}[\/tex], no intervalo [tex][0, \\pi][\/tex]. Solu\u00e7\u00e3o 1 Nesta solu\u00e7\u00e3o, usaremos uma das transforma\u00e7\u00f5es de soma em produto da trigonometria: [tex]\\cos(p) + \\cos(q) = 2 \\cdot \\cos\\left(\\dfrac{p + &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problemao-equacao-trigonometrica\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":179,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-79047","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/79047","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/users\/179"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=79047"}],"version-history":[{"count":14,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/79047\/revisions"}],"predecessor-version":[{"id":186486,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/79047\/revisions\/186486"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=79047"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}