Home / phương trình 8 cosx= căn 3/sinx+1/cosx có nghiệm là x = pi/16 +k pi /2 hay x = 4pi/npi PHƯƠNG TRÌNH 8 COSX= CĂN 3/SINX+1/COSX CÓ NGHIỆM LÀ X = PI/16 +K PI /2 HAY X = 4PI/NPI 11/11/2021 Given displaystylesinx=fracsqrt32,cosx=-frac12 , how vị you find the remaining trigonometric functions?https://socratic.org/questions/given-sinx-sqrt3-2-cosx-1-2-how-do-you-find-the-remaining-trigonometric-function displaystyle anx=-sqrt3displaystylesecx=-2displaystylecscx=frac23sqrt3displaystylecotx=-fracsqrt33 ...Bạn đang xem: Phương trình 8 cosx= căn 3/sinx+1/cosx có nghiệm là x = pi/16 +k pi /2 hay x = 4pi/npihttps://math.stackexchange.com/questions/1672613/number-of-solutions-of-8-sinx-frac-sqrt3-cosx-frac1-sinx Multiply through by sin xcos x. We get 8cos x(1-cos^2 x)=sqrt3sin x+cos x. ReHotline that cos 3x=4cos^3 x-3cos x. So the left-hand side becomes 8cos x-2(cos 3x+3cos x), that is, ...a, equals, left parenthesis, start fraction, x, divided by, 2, end fraction, right parenthesis , and 2a = xWe ..." role="text" class="MathExpression_mathExpression__22QI1 ">https://math.stackexchange.com/questions/1445550/finding-a-general-solution-for-sinx-sqrt3-2-cosx-0-5-using-an-algebra For this special case you can write: z=cos(x)+isin(x)=-frac12+ifracsqrt32=e^i(frac23pi+2pi k) chú ý that frac23pi+2pi k=-frac43pi+2pi k'. You instantly get ...Xem thêm: Đề Thi Violympic Toán Lớp 1 Vòng 15 Năm 2016 - 2017, Toán Violympic Lớp 1 Vòng 15 Năm Học 2016https://math.stackexchange.com/questions/2886859/parametrizing-the-intersection-of-a-cylinder-and-a-sphere x^2+y^2=frac14 left( x+frac12 ight)^2 + y^2 + z^2 = 1 Expvà x^2 + x + frac14 + y^2 + z^2 = 1 Collect left(x^2+y^2 ight)+x+frac14+z^2=1 Substitute first equation frac14+x+frac14+z^2=1 ... First thing first, if you make the substitution t=sin^2x the polynomial you get is -64t^3+80t^2-20t+1=0 Now, to decompose it, you could use Ruffini's rule : first we find a zero of the ...More Items left< eginarray l l 2 & 3 \ 5 và 4 endarray ight> left< eginarray l l l 2 và 0 & 3 \ -1 & 1 và 5 endarray ight>EnglishDeutschEspañolFrançaisItalianoPortuguêsРусский简体中文繁體中文Bahasa MelayuBahasa Indonesiaالعربية日本語TürkçePolskiעבריתČeštinaNederlandsMagyar Nyelv한국어SlovenčinaไทยελληνικάRomânăTiếng Việtहिन्दीঅসমীয়াবাংলাગુજરાતીಕನ್ನಡकोंकणीമലയാളംमराठीଓଡ଼ିଆਪੰਜਾਬੀதமிழ்తెలుగు
