solving Trigonometric Equations cos2x+2sin^2x-sinx=0?
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,...
895946
There is a trig identity you need: cos(2x) = 2cos^2(x)-1.\nIf you want to derive it, you can use the cosine of sum formula:\ncos(x+x) = cos(x)cos(x)-sin(x)sin(x) and go from there...\n\nYou can apply this to your equation cosx+cos(2x)=0\n\nThis gives cos(x)+2cos^2(x)-1 = 0\n\nThen yes, you can substitute p for cos(x) giving\n\n2p^2 + p - 1 = 0\n\n(2p-1)(p+1) = 0\n\np = 1/2, p = -1\n\ncos(x) = 1/2, -1\n\nx = Arccos(1/2) or Arccos(-1)\n\nx = 60 degrees or 180 degrees
935340
= cosx[cosxcosy+sinxsiny]+sinx[sinxcosy-cosxsiny]\n\n= (Cosx)^2cosy+cosxsinxsiny+(sinx)^2cosy-cosxsinxsiny\n\n= (Cosx)^2cosy + (Sinx)^2cosy\n\n= Cosy [ (cos)x^2 + (sinx)^2]\n\n= Cos(y)
413333
let suppose that \nr(@) = 4/ sqrt( 4cos^2(@) +sin^2(@) )\nthen x(@(t) ) = r(@) . cos (@) emplies\n\nx^2 (t) = r^2 (@) . cos^2(@)\n\nx^2 = 16 / (4 cos^2 + sin^2)  . cos^2\n\nso x^2(t) = 16/ (4+ tan^2(t) )\n\nBut tan( @(t) ) = y(t) / x(t)\n\nThus: x^2 = 16 / (4+ y^2 / x^2 )\nThus: \nEQ : x^2/4 + y^2/16 =  1\n\nCheck @=0  --> r(@=0) = 2\nAnd (x=2 , y=0 ) is solution of EQ
950350
LHS=sin^2x-cos^2x=(sinx+cosx)(sinx-cosx)\nRHS=1-2sinxcosx\n=(sin^2x+cos^2x)-2sinxcosx as sin^2x+cos^2x=1\n=(sinx-cosx)^2\nLHS and RHS can be equal only ifsinx+cosx=sinx-cosx\nso it is not an identity\nan identity must hole for all values
870598
(sin x+ cos x)^2 = sin^2x + 2sin x cos x + cos^2 x using (a+b)^2\n= sin^2 x + cos ^2 x + 2 sinx cos x\n= 1 + sin 2x\nbecuase sin^2x+ cos^2x =1 \nand sin 2x = sin(x+x) = sin x cos x + cosx sin x = 2sin xcos x\nusing the formula sin(A+B) = sin A cos B + cos A sin B
388285
equate the two equations: \nsin 2x = pi/4 \n2x = inverse sine pi/4 + k2pi or (pi - inverse sine pi/4) + k2pi\n2x = 0.90 + k2pi or 2.24 + k2pi\nx = 0.45 + kpi or 1.12 + kpi\nx = 0.45, 1.12, 3.59, 4.26 ( k = 0,1)
873429
cos5x = sinx\ncos5x = cos(pi/2 - x)\n5x = 2npi +- (pi/2 - x)\n6x = (4n + 1)pi/2   ,   4x= (4n-1)pi/2\nx= (4n+1)pi/12    ,  x = (4n-1)pi/8
200647
sin^2(x)=0.5*(1-cos(2x)).this can be integrated
282780
1.sin(2x)=2sin(x)cos(x)\n2.tan(x)=sin(x)/cos(x)\n3.cos(x)=(1-sin^{2}(x))^(1/2)\n\nwe can use (3) in (2) and (1) to eliminate cos(x). Then substitute sin(x)=y:\n\n1.sin(2x)=2y.(1-y^{2})^(1/2)\n2.tan(x)= y/(1-y^{2})^(1/2)\n\ntherefore,\n\nf(y)=2y(1-y^{2})^{1/2}\n     +y/(1-y^{2})^{1/2}\n\nI hope that's what you wanted.
836187
yes.\nI would solve first the square root of 1+sin2x\nsince (sinx)^2+(cosx)^2=1,so we can write\n1+sin2x=(sinx)^2+(cosx)^2+sin2x\n             =(sinx)^2+(cosx)^2+2sinx cosx             ( because we have the formula  sin2x=2sinx cosx )\n              =(sinx +cosx )^2\nso sq. root of 1+sin2x=  sinx +cosx \nnow integral of sinx+cosx is -cosx+sinx
696016
sin^2 + cos^2 = 1\nsin^2 = 1 - cos^2\n\nsub in:\n\ncos^2 - 1 + cos^2 = 0\n2*cos^2 = 1\ncos = 1/sqrt(2) or -1/sqrt(2)\n\ncos(x) = +-1/sqrt(2)\ncos(x) = +-[1/sqrt(2)]*[sqrt(2)*sqrt(2)]\ncos(x) = +-sqrt(2)/2\n\nx = 45, 135, 225, 315 deg
1150095
its sooo easy  \nsinA-sinB = 2 sin[(A-B)/2] * cos[(A-B)/2]\ncosA - cosB = 2 cos[(A-B)/2] * cos[(A-B)/2]\nthus LHS = \n(2 sin[(A-B)/2] * cos[(A-B)/2]) / (2 cos[(A-B)/2] * cos[(A-B)/2])\n= sin[(A-B)/2] / cos[(A-B)/2]\n= tan[(A-B)/2 \n= tan (A/2-B/2)\n=RHS \n\nhence proved
438454
Change cscx to 1/sinx and sin(-x) = -sinx\ncscx(cscx + sin-x)\n= (1/sinx)(1/sinx - sinx)\nGet a common denominator\n= (1/sinx)([1-(sinx)^2]/sinx)\n1-(sinx)^2 = (cosx)^2\nSo\n(1/sinx)([1-(sinx)^2]/sinx)\n= (1/sinx)((cosx)^2/sinx)\n= (cosx)^2/(sinx)^2\n= (cosx/sinx)^2\n= (cotx)^2
913797
You can begin by converting sin x^2 to 1-cos2x and since they have the same denominator x^2 just split it in two integrals (1/x^2) - (cos2x/x^2).\nThe first one is easy to solve, do initegration by parts for the second integral.
356153
using trig identities\nsin^2x+cos^2x=1, so 1-sin^2x=cos^2x and sin^2x=1-cos^2x\nsubstitue\ncos^2x(cosx)-2sin^2cosx\ncos^3x-2sin^2xcosx\ncosx(cos^2x-2*(1-cos^2x))\ncosx(cos^2x-2+2cos^2x)\ncosx(3cos^2x-2)
859883
cosx(1-2sin^2x)=0\ncosx(cos2x)=0\ncosx=0\nx=pi/2,3pi/2\ncos2x=0\n2x=pi/2 or 3pi/2\nso x=pi/4 or 3pi/4