if sin2x=a (sinx-cosx)?(how solve this problem)?
355137
first, this equation is wrong, it should look likle this:\nsin2x=2(sinx (multiplied by) cosx). If yu want, i'll try to solve the true equation for yu.
1028428
well..\n\nI'll give you one:\n\ntanx + cotx = 1/(sinx cosx)\n\ntanx = sinx/cosx\ncotx = cosx/sinx\n\nsubstituting:\n\nsinx/cosx + cosx/sinx\n\nmultiply each term to get a common denominator:\n\nsinx/cosx *(sinx/sinx) + cosx/sinx *(cosx/cosx)\n\nsin^2 x / (cosx*sinx) + cos^2x/(cosx*sinx)\n\nsince sin^2x + cos^2 x = 1, and you can add the two terms together because they have a common denominator, the entire expression equals:\n\n1/(sinxcosx)
1154463
cos(x)*cos(x) = cos^2(x)= 1-sin^2(x)\ntherefore the problem is now down to 1-sin(x)-Sin^2(x)\nthen integrate each one separately. \nfor inter of 1 is just x\nintegral of  -sin(x) is cos(x)\nand for sin^2 use the half angle identity \nsin^2(x) = .5*(1-cos(2X))\nand therefore -sin^2(x) = -.5*(1-cos(2X))\nintergral of which is -.5x+.25sin(2x)\n\nand the final answer add them and you'll get \n.5x + cos(x) + .25sin(2x)\nor \n.5x + cos(x) + .5sin(x)*cos(x)\nbecause sin(2x) = 2sin(x)*cos(x)
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
510441
Divide both sides by cosX. We get 8sinX = 4. Now divide both sides by 8, then reduce. We get sinX = 1/2.\n\nThere are two solutions when looking at the sine curve:\n\nx = pi/6\nx = (5*pi)/6\n\nits very easy, make an honest effort to understand trig, and when you do the payoffs are exponential!
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)
953609
2cos(x) - sin(2x) = 0 \nremembering that sin(2x) = 2sin(x)cos(x), \n2cos(x) -2sin(x)cos(x) = 0\n2cos(x) * [1 - sin(x)] = 0\nor cos(x) = 0 or 1-sin(x) = 0\ncos(x) = 0 or sin(x) = 1\n\nπ/2, 3π/2
200647
sin^2(x)=0.5*(1-cos(2x)).this can be integrated
240201
u dont want to prove that...u want to solve for x.  you want to find the value of x so that sin(x)=cos(x).  a possible solution would be: root three over two.  aka square root of three divided by two.  This isnt a proof....cause to be a proof it would have to read sin(x)=cos(x) which is not always true so it isnt a proof.
994111
First, factor the left hand side of the equation:\n\n2sin(x)*cos(x) + cos(x) = 0\n\ncos(x)*[2sin(x) + 1] = 0\n\nNow, this equation is satisfied whenever either cos(x) = 0 or (2sin(x)+1) = 0\n\ncos(x) = 0 at x = pi/2 and 3*pi/2  (You should memorize when the cosine and sine functions are equal to zero!)\n\nThese are the first two solutions to the original problem.\nNow deal with the factor containing the sine term.  We need to find the values of x when:\n\n2sin(x) + 1 = 0\n\nRearrange this equation to read:\n\n2sin(x) = -1\n\nthen\n\nsin(x) = -1/2\n\nx = 11pi/6 and 7pi/6 .  These are the other two solutions to the original problem.\n\nYou should also memorize when the sine and cosine functions are equal to 1/2.  For the sine function, this happens at x = pi/6 = 60 degrees and 5pi/6 = 150 degrees.  If the sine or cosine function has a certain value at x, then it will have -1 times that value at x + 180 degrees (x + pi radians), so in this case if sin(pi/6) = 1/2, then sin(pi/6 + pi) = sin(7pi/6) = -1/2.  Similarly, if sin(5pi/6) = 1/2, then sin(5pi/6 + pi) = sin(pi/6+pi) = sin(7pi/6) = -1/2.\n\nNote that despite what the previous answerer said, the sin(2pi/3) and the sin(4pi/3) do not equal -1/2.  [It's the cosine that's equal to -1/2 at these points.)
700872
d sin ax=a*cos ax dx\nd bx = b dx
699547
sin[x-(pi/2)]=kcosx\n-sin[(pi/2)-x]=kcosx\n-cosx=kcosx\nk=-1\ncos[x+{pi/2)]=ksinx\n=cos[(pi/2)+x]=ksinx\n=-sinx=ksinx\nk=-1
820147
sin(2a)=2*sin(a)*cos(a)\ncos(2a)=cos^2(a)-sin^2(a)
617038
sin(pi/2 + x) = sin (pi/2)cosx + cos(pi/2)sinx\n\n= 1cosx + 0 = cosx\n\ncos(-x)=cos(x)\n\nSo cos(x)cos(x) = cos^2(x)
640359
let sinx=u\nthen cosxdx=du\n\nand ∫[(cosx)/(9+sin^2x)]dx=\n\n∫du/(9+u^2)=        \n\n\n ( now use this formula\n∫du/(a^2+u^2)= 1/a arctan(u/a)+C)\n\n\n=1/3 arctan(u/3) +C=\n\n\n=(1/3)arctan(sinx/3)+C
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
881174
use formulas for cos(2x) and sin(2x) depending on cos(x) and sin (x) and then use the fact that\nsin(2x) = sqrt(1-cos(2x)^2).\nyou will get a quadratic equation that you solve with quadratic formula