{"id":2681,"date":"2023-02-24T21:54:18","date_gmt":"2023-02-24T20:54:18","guid":{"rendered":"http:\/\/matma.com.pl\/?p=2681"},"modified":"2023-05-06T10:01:54","modified_gmt":"2023-05-06T08:01:54","slug":"zadanie-18-0-5-zbior-rozszerzenie","status":"publish","type":"post","link":"http:\/\/matma.com.pl\/?p=2681","title":{"rendered":"zadanie 18 (0-5) zbi\u00f3r rozszerzenie"},"content":{"rendered":"<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\cos&amp;space;^2x-\\frac{2\\sqrt{3}}{3}\\sin&amp;space;x\\cos&amp;space;x-\\sin&amp;space;^2x=0\" alt=\"\\cos ^2x-\\frac{2\\sqrt{3}}{3}\\sin x\\cos x-\\sin ^2x=0\" align=\"absmiddle\" \/><\/p>\n<p>Podzielimy r\u00f3wnanie obustronnie przez <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\cos&amp;space;^{2}x\\neq&amp;space;0\" alt=\"\\cos ^{2}x\\neq 0\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\cos&amp;space;^{2}x\\neq&amp;space;0\\;&amp;space;\\;&amp;space;\\Rightarrow\\;&amp;space;\\;&amp;space;\\cos&amp;space;x\\neq&amp;space;0\\;&amp;space;\\;&amp;space;\\Rightarrow&amp;space;\\;&amp;space;\\;&amp;space;x\\neq&amp;space;\\frac{\\pi&amp;space;}{2}+k\\pi,&amp;space;\\;&amp;space;\\;&amp;space;k\\in&amp;space;\\mathbb{Z}\" alt=\"\\cos ^{2}x\\neq 0\\; \\; \\Rightarrow\\; \\; \\cos x\\neq 0\\; \\; \\Rightarrow \\; \\; x\\neq \\frac{\\pi }{2}+k\\pi, \\; \\; k\\in \\mathbb{Z}\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\cos&amp;space;^2x-\\frac{2\\sqrt{3}}{3}\\sin&amp;space;x\\cos&amp;space;x-\\sin&amp;space;^2x=0\\;&amp;space;\\;&amp;space;\/:\\cos&amp;space;^{2}x\" alt=\"\\cos ^2x-\\frac{2\\sqrt{3}}{3}\\sin x\\cos x-\\sin ^2x=0\\; \\; \/:\\cos ^{2}x\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?1-\\frac{2\\sqrt{3}}{3}\\frac{\\sin&amp;space;x}{\\cos&amp;space;x}-\\frac{\\sin&amp;space;^2x}{\\cos&amp;space;^{2}x}=0\" alt=\"1-\\frac{2\\sqrt{3}}{3}\\frac{\\sin x}{\\cos x}-\\frac{\\sin ^2x}{\\cos ^{2}x}=0\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?1-\\frac{2\\sqrt{3}}{3}\\mathrm{tg}x-\\mathrm{tg}^{2}x=0\" alt=\"1-\\frac{2\\sqrt{3}}{3}\\mathrm{tg}x-\\mathrm{tg}^{2}x=0\" align=\"absmiddle\" \/><\/p>\n<p>Pos\u0142u\u017cymy si\u0119 podstawieniem <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\mathrm{tg}x=t\" alt=\"\\mathrm{tg}x=t\" align=\"absmiddle\" \/> i <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?t\\in&amp;space;\\mathbb{R}-\\left&amp;space;\\{&amp;space;k\\pi&amp;space;\\right&amp;space;\\},\\;&amp;space;\\;&amp;space;k\\in&amp;space;\\mathbb{Z}\" alt=\"t\\in \\mathbb{R}-\\left \\{ k\\pi \\right \\},\\; \\; k\\in \\mathbb{Z}\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?1-\\frac{2\\sqrt{3}}{3}t-t^2=0\\;&amp;space;\\;&amp;space;\/\\cdot&amp;space;3\" alt=\"1-\\frac{2\\sqrt{3}}{3}t-t^2=0\\; \\; \/\\cdot 3\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?-3t^2-2\\sqrt{3}t+3=0\" alt=\"-3t^2-2\\sqrt{3}t+3=0\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\Delta&amp;space;=\\left&amp;space;(&amp;space;-2\\sqrt{3}&amp;space;\\right&amp;space;)^2-4\\cdot&amp;space;\\left&amp;space;(&amp;space;-3&amp;space;\\right&amp;space;)\\cdot&amp;space;3=4\\cdot&amp;space;3+36=12+36=48\" alt=\"\\Delta =\\left ( -2\\sqrt{3} \\right )^2-4\\cdot \\left ( -3 \\right )\\cdot 3=4\\cdot 3+36=12+36=48\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\sqrt{\\Delta&amp;space;}=\\sqrt{48}=4\\sqrt{3}\" alt=\"\\sqrt{\\Delta }=\\sqrt{48}=4\\sqrt{3}\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?t_{1}=\\frac{2\\sqrt{3}-4\\sqrt{3}}{2\\cdot&amp;space;\\left&amp;space;(&amp;space;-3&amp;space;\\right&amp;space;)}=\\frac{-2\\sqrt{3}}{-6}=\\frac{\\sqrt{3}}{3}\" alt=\"t_{1}=\\frac{2\\sqrt{3}-4\\sqrt{3}}{2\\cdot \\left ( -3 \\right )}=\\frac{-2\\sqrt{3}}{-6}=\\frac{\\sqrt{3}}{3}\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?t_{1}=\\frac{2\\sqrt{3}+4\\sqrt{3}}{2\\cdot&amp;space;\\left&amp;space;(&amp;space;-3&amp;space;\\right&amp;space;)}=\\frac{6\\sqrt{3}}{-6}=-\\sqrt{3}\" alt=\"t_{1}=\\frac{2\\sqrt{3}+4\\sqrt{3}}{2\\cdot \\left ( -3 \\right )}=\\frac{6\\sqrt{3}}{-6}=-\\sqrt{3}\" align=\"absmiddle\" \/><\/p>\n<p>Mamy zatem:<\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\mathrm{tg}x=\\frac{\\sqrt{3}}{3}\" alt=\"\\mathrm{tg}x=\\frac{\\sqrt{3}}{3}\" align=\"absmiddle\" \/> lub <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\mathrm{tg}x=-\\sqrt{3}\" alt=\"\\mathrm{tg}x=-\\sqrt{3}\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\mathrm{tg}x=-\\sqrt{3}\\;&amp;space;\\;&amp;space;\\Rightarrow&amp;space;\\;&amp;space;\\;&amp;space;x=-\\frac{\\pi&amp;space;}{3}+k\\pi&amp;space;,\\;&amp;space;\\;&amp;space;k\\in&amp;space;\\mathbb{Z}\" alt=\"\\mathrm{tg}x=-\\sqrt{3}\\; \\; \\Rightarrow \\; \\; x=-\\frac{\\pi }{3}+k\\pi ,\\; \\; k\\in \\mathbb{Z}\" align=\"absmiddle\" \/><\/p>\n<p>Uwzgl\u0119dniaj\u0105c przedzia\u0142 z tre\u015bci zadania <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?x\\in&amp;space;\\left&amp;space;[&amp;space;-\\pi&amp;space;,\\pi&amp;space;\\right&amp;space;]\" alt=\"x\\in \\left [ -\\pi ,\\pi \\right ]\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?x=-\\frac{5\\pi&amp;space;}{6}\" alt=\"x=-\\frac{5\\pi }{6}\" align=\"absmiddle\" \/> lub <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?x=\\frac{\\pi&amp;space;}{6}\" alt=\"x=\\frac{\\pi }{6}\" align=\"absmiddle\" \/> lub <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?x=-\\frac{\\pi&amp;space;}{3}\" alt=\"x=-\\frac{\\pi }{3}\" align=\"absmiddle\" \/> lub <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?x=\\frac{2\\pi&amp;space;}{3}\" alt=\"x=\\frac{2\\pi }{3}\" align=\"absmiddle\" \/><\/p>\n<p>Zajmijmy si\u0119 teraz przypadkiem, w kt\u00f3rym <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\cos&amp;space;^{2}x=&amp;space;0\\;&amp;space;\\;&amp;space;\\Rightarrow&amp;space;\\cos&amp;space;x=0\" alt=\"\\cos ^{2}x= 0\\; \\; \\Rightarrow \\cos x=0\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\cos&amp;space;x=0\\;&amp;space;\\;&amp;space;\\Rightarrow&amp;space;\\;&amp;space;\\;&amp;space;x=\\frac{\\pi&amp;space;}{2}+k\\pi&amp;space;,\\;&amp;space;\\;&amp;space;k\\in&amp;space;\\mathbb{Z}\" alt=\"\\cos x=0\\; \\; \\Rightarrow \\; \\; x=\\frac{\\pi }{2}+k\\pi ,\\; \\; k\\in \\mathbb{Z}\" align=\"absmiddle\" \/><\/p>\n<p>W\u00f3wczas:<\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\cos&amp;space;^{2}x=0\" alt=\"\\cos ^{2}x=0\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\cos&amp;space;x=0\" alt=\"\\cos x=0\" align=\"absmiddle\" \/><\/p>\n<p>Je\u015bli chodzi o funkcj\u0119 sinusa to musimy rozpatrze\u0107 dwa warianty, odpowiednio dla parzystych i nieparzystych <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?k\\in&amp;space;\\mathbb{Z}\" alt=\"k\\in \\mathbb{Z}\" align=\"absmiddle\" \/>:<\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\sin&amp;space;x=\\left\\{\\begin{matrix}&amp;space;1\\\\&amp;space;-1&amp;space;\\end{matrix}\\right.\" alt=\"\\sin x=\\left\\{\\begin{matrix} 1\\\\ -1 \\end{matrix}\\right.\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\sin&amp;space;^{2}x=1\" alt=\"\\sin ^{2}x=1\" align=\"absmiddle\" \/><\/p>\n<p>Nasze r\u00f3wnanie <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\cos&amp;space;^2x-\\frac{2\\sqrt{3}}{3}\\sin&amp;space;x\\cos&amp;space;x-\\sin&amp;space;^2x=0\" alt=\"\\cos ^2x-\\frac{2\\sqrt{3}}{3}\\sin x\\cos x-\\sin ^2x=0\" align=\"absmiddle\" \/> przyjmuje posta\u0107:<\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\left\\{\\begin{matrix}&amp;space;0-\\frac{2\\sqrt{3}}{3}\\cdot&amp;space;\\left&amp;space;(-1&amp;space;\\right&amp;space;)\\cdot&amp;space;0-\\left&amp;space;(&amp;space;-1&amp;space;\\right&amp;space;)^2=0-0-1=1\\neq&amp;space;0\\\\&amp;space;0-\\frac{2\\sqrt{3}}{3}\\cdot&amp;space;1\\cdot&amp;space;0-&amp;space;1^2=0-0-1=1\\neq&amp;space;0\\;&amp;space;\\;&amp;space;\\;&amp;space;\\;&amp;space;\\;&amp;space;\\;&amp;space;\\;&amp;space;\\;&amp;space;\\;&amp;space;\\;&amp;space;\\;&amp;space;\\end{matrix}\\right.\" alt=\"\\left\\{\\begin{matrix} 0-\\frac{2\\sqrt{3}}{3}\\cdot \\left (-1 \\right )\\cdot 0-\\left ( -1 \\right )^2=0-0-1=1\\neq 0\\\\ 0-\\frac{2\\sqrt{3}}{3}\\cdot 1\\cdot 0- 1^2=0-0-1=1\\neq 0\\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\end{matrix}\\right.\" align=\"absmiddle\" \/><\/p>\n<p>Zatem dochodzimy do sprzeczno\u015bci. Oznacza to, \u017ce <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?x=\\frac{\\pi&amp;space;}{2}+k\\pi&amp;space;,&amp;space;\\;&amp;space;\\;&amp;space;k\\in&amp;space;\\mathbb{Z}\" alt=\"x=\\frac{\\pi }{2}+k\\pi , \\; \\; k\\in \\mathbb{Z}\" align=\"absmiddle\" \/> nie spe\u0142nia naszego r\u00f3wnania.<\/p>\n<p>Ostatecznie rozwi\u0105zaniami naszego r\u00f3wnania s\u0105: <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;\\{&amp;space;-\\frac{5\\pi&amp;space;}{6},-\\frac{\\pi&amp;space;}{3},\\frac{\\pi&amp;space;}{6},\\frac{2\\pi&amp;space;}{3}&amp;space;\\right&amp;space;\\}\" alt=\"\\left \\{ -\\frac{5\\pi }{6},-\\frac{\\pi }{3},\\frac{\\pi }{6},\\frac{2\\pi }{3} \\right \\}\" align=\"absmiddle\" \/><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Podzielimy r\u00f3wnanie obustronnie przez Pos\u0142u\u017cymy si\u0119 podstawieniem i Mamy zatem: lub Uwzgl\u0119dniaj\u0105c przedzia\u0142 z tre\u015bci zadania lub lub lub Zajmijmy si\u0119 teraz przypadkiem, w kt\u00f3rym W\u00f3wczas: Je\u015bli chodzi o funkcj\u0119 sinusa to musimy rozpatrze\u0107 dwa warianty, odpowiednio dla parzystych i nieparzystych : Nasze r\u00f3wnanie przyjmuje posta\u0107: Zatem dochodzimy do sprzeczno\u015bci. Oznacza to, \u017ce nie spe\u0142nia [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-2681","post","type-post","status-publish","format-standard","hentry","category-najnowsze-pliki"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v20.1 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>zadanie 18 (0-5) zbi\u00f3r rozszerzenie - Matma.com.pl<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/matma.com.pl\/?p=2681\" \/>\n<meta property=\"og:locale\" content=\"pl_PL\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"zadanie 18 (0-5) zbi\u00f3r rozszerzenie - Matma.com.pl\" \/>\n<meta property=\"og:description\" content=\"Podzielimy r\u00f3wnanie obustronnie przez Pos\u0142u\u017cymy si\u0119 podstawieniem i Mamy zatem: lub Uwzgl\u0119dniaj\u0105c przedzia\u0142 z tre\u015bci zadania lub lub lub Zajmijmy si\u0119 teraz przypadkiem, w kt\u00f3rym W\u00f3wczas: Je\u015bli chodzi o funkcj\u0119 sinusa to musimy rozpatrze\u0107 dwa warianty, odpowiednio dla parzystych i nieparzystych : Nasze r\u00f3wnanie przyjmuje posta\u0107: Zatem dochodzimy do sprzeczno\u015bci. Oznacza to, \u017ce nie spe\u0142nia [&hellip;]\" \/>\n<meta property=\"og:url\" content=\"https:\/\/matma.com.pl\/?p=2681\" \/>\n<meta property=\"og:site_name\" content=\"Matma.com.pl\" \/>\n<meta property=\"article:published_time\" content=\"2023-02-24T20:54:18+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2023-05-06T08:01:54+00:00\" \/>\n<meta property=\"og:image\" content=\"http:\/\/latex.codecogs.com\/gif.latex?cos&amp;space;2x-frac2sqrt33sin&amp;space;xcos&amp;space;x-sin&amp;space;2x=0\" \/>\n<meta name=\"author\" content=\"lucynabartczak99\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Napisane przez\" \/>\n\t<meta name=\"twitter:data1\" content=\"lucynabartczak99\" \/>\n\t<meta name=\"twitter:label2\" content=\"Szacowany czas czytania\" \/>\n\t<meta name=\"twitter:data2\" content=\"7 minut\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/matma.com.pl\/?p=2681#article\",\"isPartOf\":{\"@id\":\"https:\/\/matma.com.pl\/?p=2681\"},\"author\":{\"name\":\"lucynabartczak99\",\"@id\":\"https:\/\/matma.com.pl\/#\/schema\/person\/b54229c3e035dea49ef8c166c5c63f1e\"},\"headline\":\"zadanie 18 (0-5) zbi\u00f3r rozszerzenie\",\"datePublished\":\"2023-02-24T20:54:18+00:00\",\"dateModified\":\"2023-05-06T08:01:54+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/matma.com.pl\/?p=2681\"},\"wordCount\":81,\"commentCount\":0,\"publisher\":{\"@id\":\"https:\/\/matma.com.pl\/#organization\"},\"articleSection\":[\"najnowsze pliki\"],\"inLanguage\":\"pl-PL\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"https:\/\/matma.com.pl\/?p=2681#respond\"]}]},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/matma.com.pl\/?p=2681\",\"url\":\"https:\/\/matma.com.pl\/?p=2681\",\"name\":\"zadanie 18 (0-5) zbi\u00f3r rozszerzenie - Matma.com.pl\",\"isPartOf\":{\"@id\":\"https:\/\/matma.com.pl\/#website\"},\"datePublished\":\"2023-02-24T20:54:18+00:00\",\"dateModified\":\"2023-05-06T08:01:54+00:00\",\"breadcrumb\":{\"@id\":\"https:\/\/matma.com.pl\/?p=2681#breadcrumb\"},\"inLanguage\":\"pl-PL\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/matma.com.pl\/?p=2681\"]}]},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/matma.com.pl\/?p=2681#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\/\/matma.com.pl\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"zadanie 18 (0-5) zbi\u00f3r rozszerzenie\"}]},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/matma.com.pl\/#website\",\"url\":\"https:\/\/matma.com.pl\/\",\"name\":\"Matma.com.pl\",\"description\":\"\",\"publisher\":{\"@id\":\"https:\/\/matma.com.pl\/#organization\"},\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/matma.com.pl\/?s={search_term_string}\"},\"query-input\":\"required name=search_term_string\"}],\"inLanguage\":\"pl-PL\"},{\"@type\":\"Organization\",\"@id\":\"https:\/\/matma.com.pl\/#organization\",\"name\":\"Matma.com.pl\",\"url\":\"https:\/\/matma.com.pl\/\",\"logo\":{\"@type\":\"ImageObject\",\"inLanguage\":\"pl-PL\",\"@id\":\"https:\/\/matma.com.pl\/#\/schema\/logo\/image\/\",\"url\":\"http:\/\/matma.com.pl\/wp-content\/uploads\/2020\/07\/cropped-trudne-1.jpg\",\"contentUrl\":\"http:\/\/matma.com.pl\/wp-content\/uploads\/2020\/07\/cropped-trudne-1.jpg\",\"width\":150,\"height\":30,\"caption\":\"Matma.com.pl\"},\"image\":{\"@id\":\"https:\/\/matma.com.pl\/#\/schema\/logo\/image\/\"}},{\"@type\":\"Person\",\"@id\":\"https:\/\/matma.com.pl\/#\/schema\/person\/b54229c3e035dea49ef8c166c5c63f1e\",\"name\":\"lucynabartczak99\",\"image\":{\"@type\":\"ImageObject\",\"inLanguage\":\"pl-PL\",\"@id\":\"https:\/\/matma.com.pl\/#\/schema\/person\/image\/\",\"url\":\"https:\/\/secure.gravatar.com\/avatar\/a77fc36c9912d50deeefdd4c81a75615386fbe2bdaccb30be7a599c7e140e577?s=96&d=mm&r=g\",\"contentUrl\":\"https:\/\/secure.gravatar.com\/avatar\/a77fc36c9912d50deeefdd4c81a75615386fbe2bdaccb30be7a599c7e140e577?s=96&d=mm&r=g\",\"caption\":\"lucynabartczak99\"},\"url\":\"http:\/\/matma.com.pl\/?author=1\"}]}<\/script>\n<!-- \/ Yoast SEO plugin. -->","yoast_head_json":{"title":"zadanie 18 (0-5) zbi\u00f3r rozszerzenie - Matma.com.pl","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/matma.com.pl\/?p=2681","og_locale":"pl_PL","og_type":"article","og_title":"zadanie 18 (0-5) zbi\u00f3r rozszerzenie - Matma.com.pl","og_description":"Podzielimy r\u00f3wnanie obustronnie przez Pos\u0142u\u017cymy si\u0119 podstawieniem i Mamy zatem: lub Uwzgl\u0119dniaj\u0105c przedzia\u0142 z tre\u015bci zadania lub lub lub Zajmijmy si\u0119 teraz przypadkiem, w kt\u00f3rym W\u00f3wczas: Je\u015bli chodzi o funkcj\u0119 sinusa to musimy rozpatrze\u0107 dwa warianty, odpowiednio dla parzystych i nieparzystych : Nasze r\u00f3wnanie przyjmuje posta\u0107: Zatem dochodzimy do sprzeczno\u015bci. Oznacza to, \u017ce nie spe\u0142nia [&hellip;]","og_url":"https:\/\/matma.com.pl\/?p=2681","og_site_name":"Matma.com.pl","article_published_time":"2023-02-24T20:54:18+00:00","article_modified_time":"2023-05-06T08:01:54+00:00","og_image":[{"url":"http:\/\/latex.codecogs.com\/gif.latex?\\cos&amp;space;^2x-\\frac{2\\sqrt{3}}{3}\\sin&amp;space;x\\cos&amp;space;x-\\sin&amp;space;^2x=0"}],"author":"lucynabartczak99","twitter_card":"summary_large_image","twitter_misc":{"Napisane przez":"lucynabartczak99","Szacowany czas czytania":"7 minut"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/matma.com.pl\/?p=2681#article","isPartOf":{"@id":"https:\/\/matma.com.pl\/?p=2681"},"author":{"name":"lucynabartczak99","@id":"https:\/\/matma.com.pl\/#\/schema\/person\/b54229c3e035dea49ef8c166c5c63f1e"},"headline":"zadanie 18 (0-5) zbi\u00f3r rozszerzenie","datePublished":"2023-02-24T20:54:18+00:00","dateModified":"2023-05-06T08:01:54+00:00","mainEntityOfPage":{"@id":"https:\/\/matma.com.pl\/?p=2681"},"wordCount":81,"commentCount":0,"publisher":{"@id":"https:\/\/matma.com.pl\/#organization"},"articleSection":["najnowsze pliki"],"inLanguage":"pl-PL","potentialAction":[{"@type":"CommentAction","name":"Comment","target":["https:\/\/matma.com.pl\/?p=2681#respond"]}]},{"@type":"WebPage","@id":"https:\/\/matma.com.pl\/?p=2681","url":"https:\/\/matma.com.pl\/?p=2681","name":"zadanie 18 (0-5) zbi\u00f3r rozszerzenie - Matma.com.pl","isPartOf":{"@id":"https:\/\/matma.com.pl\/#website"},"datePublished":"2023-02-24T20:54:18+00:00","dateModified":"2023-05-06T08:01:54+00:00","breadcrumb":{"@id":"https:\/\/matma.com.pl\/?p=2681#breadcrumb"},"inLanguage":"pl-PL","potentialAction":[{"@type":"ReadAction","target":["https:\/\/matma.com.pl\/?p=2681"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/matma.com.pl\/?p=2681#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/matma.com.pl\/"},{"@type":"ListItem","position":2,"name":"zadanie 18 (0-5) zbi\u00f3r rozszerzenie"}]},{"@type":"WebSite","@id":"https:\/\/matma.com.pl\/#website","url":"https:\/\/matma.com.pl\/","name":"Matma.com.pl","description":"","publisher":{"@id":"https:\/\/matma.com.pl\/#organization"},"potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/matma.com.pl\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"pl-PL"},{"@type":"Organization","@id":"https:\/\/matma.com.pl\/#organization","name":"Matma.com.pl","url":"https:\/\/matma.com.pl\/","logo":{"@type":"ImageObject","inLanguage":"pl-PL","@id":"https:\/\/matma.com.pl\/#\/schema\/logo\/image\/","url":"http:\/\/matma.com.pl\/wp-content\/uploads\/2020\/07\/cropped-trudne-1.jpg","contentUrl":"http:\/\/matma.com.pl\/wp-content\/uploads\/2020\/07\/cropped-trudne-1.jpg","width":150,"height":30,"caption":"Matma.com.pl"},"image":{"@id":"https:\/\/matma.com.pl\/#\/schema\/logo\/image\/"}},{"@type":"Person","@id":"https:\/\/matma.com.pl\/#\/schema\/person\/b54229c3e035dea49ef8c166c5c63f1e","name":"lucynabartczak99","image":{"@type":"ImageObject","inLanguage":"pl-PL","@id":"https:\/\/matma.com.pl\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/a77fc36c9912d50deeefdd4c81a75615386fbe2bdaccb30be7a599c7e140e577?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/a77fc36c9912d50deeefdd4c81a75615386fbe2bdaccb30be7a599c7e140e577?s=96&d=mm&r=g","caption":"lucynabartczak99"},"url":"http:\/\/matma.com.pl\/?author=1"}]}},"_links":{"self":[{"href":"http:\/\/matma.com.pl\/index.php?rest_route=\/wp\/v2\/posts\/2681","targetHints":{"allow":["GET"]}}],"collection":[{"href":"http:\/\/matma.com.pl\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"http:\/\/matma.com.pl\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"http:\/\/matma.com.pl\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"http:\/\/matma.com.pl\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=2681"}],"version-history":[{"count":2,"href":"http:\/\/matma.com.pl\/index.php?rest_route=\/wp\/v2\/posts\/2681\/revisions"}],"predecessor-version":[{"id":2683,"href":"http:\/\/matma.com.pl\/index.php?rest_route=\/wp\/v2\/posts\/2681\/revisions\/2683"}],"wp:attachment":[{"href":"http:\/\/matma.com.pl\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=2681"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"http:\/\/matma.com.pl\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=2681"},{"taxonomy":"post_tag","embeddable":true,"href":"http:\/\/matma.com.pl\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=2681"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}