cos squared 2x + 4sin squared x times cos squared x = 1?
411302
cos^2 (2x) is equal to 1-sin^2(2x), remember, sin^2(x)+cos^2(x)=1
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)
276083
Yes, it can be rewritten as 2sin2xcos2x.  Or, you can use the addition formula for sineto prove this.  sin(x + y) = sinxcosy + cosxsiny.  In your case, sin(2x + 2x) = sin2xcos2x + cos2xsin2x = 2sin2xcos2x.
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
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
336698
2. (2sin^2 x - 1)^2 / sin^4 x - cos^4 x = 1 - 2cos^2 x\n\nAssuming you meant\n(2sin^2 x - 1)^2 / (sin^4 x - cos^4 x)\nthen\n\n(2sin^2 x - 1)^2 / (sin^4 x - cos^4 x)\n = (2sin^2 x - 1)^2 / ((sin^2 x + cos^2 x)(sin^2 x - cos^2 x))\n = (2sin^2 x - 1)^2 / (sin^2 x - cos^2 x)\n = (2(1 - cos^2 x) - 1)^2 / ((1 - cos^2 x) - cos^2)\n = (1 - 2cos^2x)^2 / (1 - 2cos^2 x)\n = 1 - 2cos^2 x\n\n\n\nUpdate (added after wiildmunkii's answer)\n= = = = = = = = = = = = = = = = = = = = =\n\n1. sin^2 x + sin^4 x / cos^2 x + cos^4 x = tan 3x\n\nIf you meant\n(sin(2x) + sin(4x)) / (cos(2x) + cos(4x)) = tan3x\nthen yes.\n\nCombining\n\nsin(a+b) = sin(a) cos(b) + cos(a) sin(b)\ncos(a+b) = cos(a) cos(b) - sin(a) sin(b)\ntan(a + b) = sin(a + b) / cos(a + b)\n = (sin(a) cos(b) + cos(a) sin(b)) / (cos(a) cos(b) - sin(a) sin(b))\n = (tan(a) + tan(b)) / (1 - tan(a) tan(b))\n\nyou get\n\nsin(2x) = sin(x) cos(x) + cos(x) sin(x)\n = 2 sin(x) cos(x)\ncos(2x) = cos(x) cos(x) - sin(x) sin(x)\n = cos^2(x) - sin^2(x)\n = cos^2(x) - (1 - cos^2(x))\n = 2 cos^2(x) - 1\n [similarly = 1 - 2 sin^2(x)]\ntan(2x) = (tan(x) + tan(x)) / (1 - tan(x) tan(x))\n = 2 tan(x) / (1 - tan^2(x))\n\n\ne^it = (cos(t) + i sin(t))\n\nSo (e^it)^n = (cos(t) + i sin(t))^n\n\nBut (e^it)^n = (e^int) = cos(nt) + i sin(nt)\n\nSo cos(nt) + i sin(nt) = (cos(t) + i sin(t))^n\n\nFor any given n, expand the right hand side using the binomial theorem and equate real and imaginary parts to get:\n\ncos(2t) + i sin(2t) = cos^2(t) - sin^2(t) + 2 i sin(t) cos(t)\ncos(2t) = cos^2(t) - sin^2(t)\n = 2 cos^2(t) - 1\n = 1 - 2 sin^2(t)\nsin(2t) = 2 sin(t) cos(t)\n\ncos(3t) + i sin(3t) =\n cos^3(t) + i (3 cos^2(t) sin(t)) - (3 cos(t) sin^2(t)) - i sin^3(t)\ncos(3t) = cos^3(t) - 3 cos(t) sin^2(t)\n = 4 cos^3(t) - 3 cos(t)\nsin(3t) = 3 cos^2(t) sin(t) - sin^3(t)\n = 3 sin(t) - 4 sin^3(t)\n\ncos(4t) + i sin(4t) =\n cos^4(t) + i (4 cos^3(t) sin(t)) - (6 cos^2(t) sin^2(t)) - i (4 cos(t) sin^3(t)) + sin^4(t)\ncos(4t) = cos^4(t) - 6 cos^2(t) sin^2(t) + sin^4(t)\n = cos^4(t) - 6 cos^2(t) (1 - cos^2(t) + (1 - cos^2(t))^2\n = 8 cos^4(t) - 8 cos^2(t) + 1\nsin(4t) = 4 cos^3(t) sin(t) - 4 cos(t) sin^3(t)\n = 4 cos(t) sin(t) (cos^2(t) - sin^2(t))\n = 4 cos(t) sin(t) (2cos^2(t) - 1)\n = 4 cos(t) sin(t) (1 - 2 sin^2(t))\n\nFinally...\n\n(sin(2x) + sin(4x)) / (cos(2x) + cos(4x))\n = (2 sin(x) cos(x) + 4 cos(x) sin(x) (1 - 2 sin^2(x))) / (2 cos^2(x) - 1 + 8 cos^4(x) - 8 cos^2(x) + 1)\n = (6 sin(x) cos(x) - 8 sin^3(x) cos(x)) / (8 cos^4(x) - 6 cos^2(x))\n # divide by 2 cos(x)\n = (3 sin(x) - 4 sin^3(x)) / (4 cos^3(x) - 3 cos^(x))\n = sin(3x) / cos(3x)\n = tan(3x)
897582
cos²x - 1 - cosx = 0\nie cos²x - cosx - 1 = 0\ncosx = ½(1 ± √5) by quadratic formula\nNow as |cosx| ≤ 1 for real x\ncosx = ½(1 - √5) = -0.61803......\nSo x = 180° - 51°49'38", 180° + 51°49'38"\n= 128°10'22", 231°49'38" (for 0° ≤ x ≤ 360°)
175133
2x-1(xsquared-x+10)squared
998890
If you mean to solve it, it would be this:\nF:xtimesx=x^2\nO:xtime6.5\nI:xtimes-4.5\nL:-4.5times6.5\nEquals:x^2+2x-29.25
971617
y = sin(3x) + cos(4x)\ny = sin(2x + x) + cos(2x + 2x)\n\nsin(2x + x) = sin(2x)cos(x) + sin(x)cos(2x)\ncos(2x + 2x) = cos(2x)cos(2x) - sin(2x)sin(2x)\n\nsin(2x) = 2sinxcosx\ncos(2x) = cos(x)^2 - sin(x)^2 = 2cos(x)^2 - 1 = 1 - 2sin(x)^2\n\n(sin(2x)cos(x) + sin(x)cos(2x)) + (cos(2x)^2 - sin(2x)^2)\n\n(2sin(x)cos(x)cos(x) + sin(x)(2cos(x)^2 - 1)) + ((2cos(x)^2 - 1)^2 - (2sin(x)cos(x))^2)\n\n(2sin(x)cos(x)^2 + 2sin(x)cos(x)^2 - sin(x)) + (((2cos(x)^2 - 1)(2cos(x)^2 - 1)) - 4sin(x)^2cos(x)^2)\n\n(4sin(x)cos(x)^2 - sin(x)) + ((4cos(x)^4 - 2cos(x)^2 - 2cos(x)^2 + 1) - 4sin(x)^2cos(x)^2)\n\n4sin(x)cos(x)^2 - sin(x) + (4cos(x)^4 - 4cos(x)^2 + 1 - 4sin(x)^2cos(x)^2)\n\n4sin(x)cos(x)^2 - sin(x) + 4cos(x)^4 - 4cos(x)^2 + 1 - 4sin(x)^2cos(x)^2\n\ncos(x)^2 = 1 - sin(x)^2\n\n4sin(x)(1 - sin(x)^2) - sin(x) + 4(1 - sin(x)^2)^2 - 4(1 - sin(x)^2) + 1 - 4sin(x)^2(1 - sin(x)^2)\n\n4sin(x) - 4sin(x)^3 - sin(x) + 4((1 - sin(x)^2)(1 - sin(x)^2)) - 4 + 4sin(x)^2 + 1 - 4sin(x)^2 + 4sin(x)^4\n\n4sin(x) - 4sin(x)^3 - sin(x) + 4(1 - sin(x)^2 - sin(x)^2 + sin(x)^4) - 4 + 4sin(x)^2 + 1 - 4sin(x)^2 + 4sin(x)^4\n\n4sin(x) - 4sin(x)^3 - sin(x) + 4(1 - 2sin(x)^2 + sin(x)^4) - 4 + 4sin(x)^2 + 1 - 4sin(x)^2 + 4sin(x)^4\n\n4sin(x) - 4sin(x)^3 - sin(x) + 4 - 8sin(x)^2 + 4sin(x)^4 - 4 + 4sin(x)^2 + 1 - 4sin(x)^2 + 4sin(x)^4\n\n(4 + 4)sin(x)^4 - 4sin(x)^3 + (-8 + 4 - 4)sin(x)^2 + (4 - 1)sin(x) + (4 - 4 + 1)\n\nsin(3x) + cos(4x) = 8sin(x)^4 - 4sin(x)^3 - 8sin(x)^2 + 3sin(x) + 1\n\nfor a graph, go to http://www.calculator.com/calcs/GCalc.html\n\njust type in sin(3x) + cos(4x), click enter
216863
((4 * 4) / 4) - 4 + 4 = 4
1155111
xsquare=x+x+x...x terms\nby diff x\n2x=1+1+1...x terms\n2x=x\n2=1
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
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
206636
for (i=1,t=1,cosx=1;i<=n;++i)\n{\n  t=(-1)*(t*x*x)/((2*i)*(2*i-1);\n  cosx += t;\n}
681236
Multiply each term in the first () by each term in the second ():\n\n2x times 2x = 4x^2\n2x times 1 = 2x\n-3 times 2x = -6x\n-3 times 1 = -3\n\nThen combine 2x and -6x\nAnswer: 4x^2 - 4x - 3
797886
Have you learned trig identities yet?  I'd check to see if there was a trig identity you could substitute here.\n\nBTW, I am assuming you meant cos²x rather than cosx².\n\nTry this:  cos²x + sin²x = 1, so substitute 1-sin²x for cos²x:\n\n6cos²x + sinx - 4 = 0\n\n6(1-sin²x) + sinx - 4 = 0\n\n-sin²x + sinx + 2 = 0\n\n(multiply by -1 to make it a little easier)\n\nsin²x - sinx - 2 = 0\n\n(sinx-2)(sinx+1) = 0\n\nSo, sinx = 2 or sinx = -1\n\nSolving for x, this means x = sin^-1(2) or sin^-1(-1)