Solve the equation sin^5(x)+cos^3(x)=1?
926243
sin^5(x) + cos³(x) = 1\n\nBy Quickmath (ref 1)\nx = 0, π/2 and 2π and a number of imaginery solutions \n\nYou can also solve it by the t formulae \nie if t = tanx/2 then sin x = 2t/(1 + t²) and cos x = (1 - t²)/(1 + t²)\n\nAfter some tedious algebra you get \n2t^2(t - 1)( polynomial of degree 7 in t) = o\n\nWhence t = 0, 0, 1 and some other irrational value.and 6 imaginery values\n\nSo x/2 = 0, π/4, π (in range 0 ≤ x/2 ≤ π)\nSo x = 0, π/2, 2π and the mystery value
733063
5x^2-35x+3x-21\n5x^2-32x-21
894390
2sin3x*[-15sin3x]\n=-30*sin^23x
546258
cos2x=cos(x+x)=cos^2x-sin^2x=1-sin^2x-sin^2x=1-2sin^2x\ncos2x=1-2sin^2x  \ncos2x+2sin^2x-sinx=1-2sin^2x+2sin^2x-sinx=1-sinx=0\nsinx=1\nx=2kπ+π/2  ,   k=...,-1,0,1,...
225231
sinx = a     with    a Є [-1;1]\n\n2sin^2 x - 5sinx + 2 = 0 \n<=> 2a^2 - 5a + 2 = 0\nWe have 2 roots: a = 2 (reject) & a= 1/2 because a Є [-1;1]\n => sinx = 1/2\n<=> x = ∏/3 + k2∏ and x = 2∏/3 + t2∏ with k,t Є Z
1073595
(cosx)^2 - 2sinx = 2\ncosx^2 = 1 - sinx^2\n(1 - sinx^2) - 2sinx = 2\n1 - sinx^2 - 2sinx = 2\n-sinx^2 - 2sinx + 1 = 2\n-sinx^2 - 2sinx - 1 = 0\n-(sinx^2 + 2sinx + 1) = 0\n-(sinx + 1)(sinx + 1)\n\n-sinx - 1 = 0\n-sinx = 1\nsinx = -1\nx = sin^-1(-1)\nx = 270°\n\n--------------------------------------\n\n1.5^2 = 1.7^2 + 1.8^2 - 2(1.7)(1.8)cosA\n2.25 = 2.89 + 3.24 - 6.12cosA\n2.25 = 6.13 - 6.12cosA\n-6.12cosA = -3.88\ncosA = (388/612)\ncosA = (97/153)\nA = 50.6551127\n\n1.7^2 = 1.5^2 + 1.8^2 - 2(1.5)(1.8)cosB\n2.89 = 2.25 + 3.24 - 5.4cosB\n2.89 = 5.49 - 5.4cosB\n-2.6 = -5.4cosB\ncosB = (13/27)\nB = 61.2177953\n\n1.8^2 = 1.5^2 + 1.7^2 - 2(1.5)(1.7)cos(C)\n3.24 =  2.25 + 2.89 - 5.1cosC\n3.24 = 5.14 - 5.1cosC\n-1.9 = -5.1cosC\ncosC = (19/51)\nC = 68.1270919\n\nAs you can see, the angle opposite of side A, is the smallest angle.\n\nANS : about 51°\n\n----------------------------------------------\n\ncos(A) = 4/7\nx = 4\nr = 7\n\nr = sqrt(x^2 + y^2)\n7 = sqrt(4^2 + y^2)\n49 = 16 + y^2\ny^2 = 33\ny = sqrt(33)\n\nsin(A) = y/r\nsinA = (-sqrt(33))/7\n\ntanA = y/x\ntanA = (-sqrt(33))/4\n\ncscA = r/y\ncscA = -7/(sqrt(33))\ncscA = (-7sqrt(33))/33\n\nsecA = r/x\nsecA = 7/4\n\ncotA = x/y\ncotA = -4/(sqrt(33))\ncotA = (-4sqrt(33))/33
64064
put, for example: t=cos x, then sin^2(x)= 1-cos^2(x)= 1-t^2\n\nThus you get from the original equation, a quadratic equation:\n\nt-2(1-t^2)+1=0 or t-2+2t^2+1=0, or\n\n2t^2+t-1 = 0.\n\nAfter factorization of the left side \n\n(2t-1)(t+1)=0\n\nthen 2t-1=0 gives one solution for t;\n\nt+1= 0 gives another solution.\n\nSince this is quadratic equation, you have received all the possible solutions.\n\nT_1 = 1/2 ;    T_2 = -1\n\nBut we have assumed earlier, to get to that equation, that\n\nT = cos (x)\n\nSo now we have to consider both options for t;\n\n1)cos x = 1/2; 2)cos x = -1\n\nFrom it goes something like that, check those calculations for yourself of course:\n\nSolution of the first is x = PI/2+ PIK, K= 0, 1..\n\nSolution of the second is x= PI+ PIK, K=0,1...\n\nThat is how you turn a trig problem into an algebraic one.\n\nHope that helps.
542056
Start by factoring:\n\n(cos^2(x))(sin^2(x))-(cos^2(x)) = (cos^2(x))(sin^2(x) - 1)\n= -(cos^2(x))(1 - sin^2(x))\n= -(cos^2(x))(cos^2(x))\n= -cos^4(x)
940941
sin 3A = sin (2A + A)\n\n= sin 2A cos A + sinA cos 2A\n\n= (2 sin A cos A) cos A + sin A (1 - 2 sin^2A)\n\n- 2 sin A cos^2 A + sin A - 2 sin^3 A\n\n= 2 sin A (1 - sin^2 A ) + sin A - 2 sin^3 A\n\n= 2 sin A - 2 sin^3 A + sin A - 2 sin^3 A\n\n= 3 sin A - 4 sin^3 A
907452
This is derived quickest from de Moivre's Thm, which is \n\ncos(nx) + isin(nx) = (cosx + isinx)^n\n\n(where i^2 = -1)\n\nExpand the RHS, separate the Re and Im terms to get both cos(nx) and sin(nx) as polynomials in cos(x) and sin(x). Your case is n=3.\n\nOr you can plough through cos(x+x+x) applying f-las for cos(a+b) and sin(a+b).
329959
1-CosA = 2 sin^2 A/2     -----1\n\nFirst whe have to find the value of the cos \n By Cos A = Root(1-Sin^2A)\nCos A = 5/13\nAnd \nThe apply the formaula of the 1\n1- 5/13 = 2Sin^2 A/2\nSinA/2 = Root(8/26) it gives the answer\n And the value of the cos A/2\nIs get by equation 1+ cos A = 2 Cos ^2 A/2
540710
x=5*sin20=1.7\n\ny=5*cos20=4.7
832532
dy/dx=cos x\n\nIntegrating both sides, we will get:\ny = sinx + c\n\ny(0) = sin0 + c = 0 => c=0\n\nSo y=sinx.
876211
2 cos^2 x + sin^2 x = 1\n\ncos^2 x + cos^2 x + sin^2 x = 1\n\ncos^2 x + (cos^2 x + sin^2 x) = 1\n\ncos^2 x + 1 = 1\n\ncos^2 x = 0\n\ncos x = 0\n\nx = pi/2 or 3 pi/2
842198
I think you mean\n\n(sin^4 t - cos^4 t )/ sin 2t where t is for O\n= (sin ^2 t+ cos^2t)(sin^2 - cos ^ 2 t) as (a+b)(a-b) = a^2 - b^2\n= 1. (- cos 2t)/sin 2t  using sin^2t + cos^2 t = 1 and cos^2 t- sin^2 t = cos 2t\n= - cot 2t
904134
sin3x = sin (x+ 2x) \n          = sinx.cos2x + cosx.sin2x\n          = sinx (2cos^2x -1) + cosx (2sinx.cosx)\n          = 2sinx.cos^2x - sinx + 2sinx.cos^2x\n          = sinx (2cos^2x - 1 + 2cos^2x)\n          = sinx (4cos^2x - 1)\n \n But, cos^2x = 1 - sin^2x\nso then,\n\n   sin3x = sinx [ 4 - 4sin^2x - 1]\n             = sinx (3 - 4sin^2x)\n\n Therefore,\n\n  sin3x/sinx =  sinx(3 - 4sin^2x)/sinx\n                   =  3 - 4sin^2x  --- answer.
842017
let - 5+ 5 sqrt(3) i = r cos t + ir sin t\n\nr = sqrt(5^2+(5sqrt(3))^2) = sqrt(25+75) = 10\n\ncos t = -1/2\n\nt = 120 degrees as cos -ve and sin +ve\n\nt = 2pi/3\n\nso C is the ans