4cos^2(2x)-4cos(2x)+1=0?
475755
let cos (2x) = y. this forms a normal quadratic equation that you can solve: \n4y^2-4y +1 = 0\n(2y-1)^2=0\n2y-1=0\ny=1/2\ncos 2x = 1/2\n2x = +/- 60 + k360\nx = +/- 30 + k180\nnow sub in values of k (k=0,1,2 etc) depending on the domain given for your solutions\nx = -30, 30, 150, 210 etc
812732
2^(4x) - 4(2^(2x)) - 12 = 0\n\nFirst, note that 2^(4x) is equal to (2^(2x))^2\n\n(2^(2x))^2 - 4(2^(2x)) - 12 = 0\n\nThen, solve it like a quadratic.  Let y = 2^2x and you get:\n\ny^2 - 4y - 12 = 0\n(y - 6) (y + 2) = 0\n\ny = 6\ny = -2\n\nSubstitute y = 2^(2x) back to get\n\n2^(2x) = 6\n2^(2x) = -2\n\nYou can throw away the second equation because you can neve get a negative result from 2^(2x).  So we just deal with the first one.\n\n2^(2x) = 6\nConvert to logarithmic form\n\nlog[base 2](6) = 2x\nx = log[base 2](6) / 2\n\nThat can be simplified even more, but I'll leave that assignment up to you.
795154
2CO(g) + O2(g) --> 2CO2(g)
757614
4 (x^2 + 3X -4)\n4(x+4)(X-1)
855361
2x^2 - 9x - 5=0\n(2x+1)(x-5)=0\n2x+1=0\nx=-1/2\nx-5=0\nx=5\n\nroots are \nx=5, -1/2
364111
take the first derivative and set it equal to zero and solve.  2x +1 = 0 x= -1/2  solve for y.  y=-13/4\nvertex is at x= -13/4 y=-2
862692
y" + y = 1\n\n1=sin(x) + xcos(x)\n\nEquation should read 1=sin^2(x) + xcos^2(x)\nand x value is elements of real numbers.
412551
-4x+16=-1-4x\n-4x+4x=-16-1\n0=-17\n\nthe answer is [a]0
788468
let a be cos2x  | a | <= 1\n4a^2 - 4a + 1 = 0\n<=> (2a - 1)^2 = 0\n<=> 2a -1 = 0 \n<=> a=1/2\n<=> cos2x = cos (pi/3)\n<=> 2x = +- pi/3 + k2pi (k is a subset of Z )\n<=> x = +- pi/6 + kpi\naccording to the problem, we have these following roots :\nx = pi/6 or x = 7pi/6 or x = -pi/6  or x = 5pi/6
896553
4x^2+8x-x-2\n(4x-1)(x+2)
940135
Use factoring.\nx2-2x-8=0\n(x-4)(x+2)=0\nso\nx-4=0 and x+2=0\nsolve\nx=4 and x=-2
411302
cos^2 (2x) is equal to 1-sin^2(2x), remember, sin^2(x)+cos^2(x)=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
268383
2x^2+6x-1x-3\n2x(1x+3)-1(1x+3)=0\n(2x-1)(x+3)=0\nx=1/2 or -3
898033
cos2x = 1 - 2 (sin x) ^2 = sin x\n\nthen 2 ( sin x)^2 + sin x  - 1 = 0\n or (2sin x - 1 )(sin x + 1) = 0\nsin x = 1/2 , - 1\nx = 30 deg, or - 90 deg
1135417
u can factor it out to be (2sinx+1)(cosx)=0.. then find x for which either sinx=-1/2 or cosx=0\n\n(sinx+1)cosx=0 same method applies
707648
Is this your equation?\nx - 2 - 2 - 3x = 0\n\n-2x - 4 = 0\n-2x = 4\nx = 4/-2\nx = -2\n\nThe answer is:\n x can either equal *-2* or *you made a mistake*