Canany one solve this equation?  y = arcTan(sqrt(3)/3)?
443794
30 degrees or pi/6 radians
704523
(sqrt(3150^2+2430^2))/15\n265.2244 m/s\n\narctan(y/x)\narctan(-2430/3150)\n-37.647 degrees
274244
Put Y=tan x and solve the equation for Y.\nThis gives two solutions Y=tan x = 1,7808 and Y =tan x = -0,2808.\nThis leads to x = 60.68 degrees and x = -15.68 degrees.\n\nAdd 360 degrees to the second solution and transform to degrees and minutes for the final answer.
932187
y = 3/2 (-4) -2 = -12/2 - 2 = -6 -2 =  -8\n\n(x,y) = (4,-8)
310146
i think u can do this:\n               \n            (secA+tanA)2\n   =sec2A-2tanA.secA+tan2A\n  now,   put the value of secA that can b derived from the question\n             secA+tanA=2\n       or,secA=2-tanA\n                =sec2A-2tanA.2-tanA+tan2A\n           =sec2A-4tanA-2tan2A+tan2A\n             =sec2A-4tanA-tan2A\n              =tan2A-1-4tanA-tan2A\n             =-(1+4tanA)\n\nthis may b the ans(in the above solution in some places 2 means square,i hope u'll understand it)
446101
switch the y and x\n\nx = 2y - 3\n\nThen solve for y\n\nx + 3 = 2y\n\n(x+3)/2 = y\n\nnow you have the inverse.
407911
So you are doing some numerical analysis... if you have a spreadsheet program you can use that. Use the trapezoid rule or Simpson's rule for some small increment in x.  Your spreadsheet program is able to evaluate arctan.  The second question can be evaluated with Newton's method.
931616
COSx = -4/5\nx = ARCCOS(-4/5)\n\nThen just plug in the x value to the rest of the equations.
502102
well chinaren got it started but I believe his explianation is weird.\nLets assume x=3^t\nthen log3(x)+log2(x+1)=3\n=>log3(3^t)+log2(3^t+1)=3\n=>tlog3(3)+log2(3^t+1)=3\n=>log2(3^t+1)=3-t\n=>3^t+1=2^(3-t)\n=>3^t+1=2^3/2^t\n=>(3^t+1)*2^t=2^3\n=>3^t*2^t+2^t=8\n=>6^t+2^t=8\nfrom here there is one one solution possible as t=1\nso x=3^1=3
966696
Arc=r.theta=8*pi/8=pi
346140
Well there are the obvious values that are known of arctan. But if your constant K doesn't fall into these categories I'm not sure how you would do this without taylor expansion.\n\nTry this link, it gives the well known values on tan(x) so reading it backwards gives the arctan values
912572
sin^3 β + cos^3 β  = (sinβ + cosβ)(sin^2β + cos^2β - sinβcosβ)\n= (sinβ + cosβ)(1 - sinβcosβ)\n\nsin^3 β + cos^3 β / sin β + cos β\n = (sinβ + cosβ)(1 - sinβcosβ)/( sin β + cos β)\n= 1 - sinβcosβ
838286
well Honestly dear Megan i don't think its correct , coz equation should have ONE " = " \n\nit must be something like this \n-4r ² ± 21r=r+13\n\nGood Luck
747583
y= √(1/(x² + 1) – 2) \n\nWell 1/(x² + 1) – 2 ≥ 0\nie 1/(x² + 1) ≥ 2\nSince x² + 1 ≥ 0 for all x\nOn taking reciprocals\nx² + 1 ≤ ½\nx² ≤ -½ This is not possible as x² ≥ 0 for all x \nSo there is no domain\n----------------------- -----------------------\nWell that makes sense as  x² + 1 ≥ 1\nSo 1/(x² + 1) ≤ 1\nSo 1/(x² + 1) - 2 ≤ -1\n\nSo taking square roots gives no real solutions.
796433
y = 3x²\n\nS(x, y) = (3x, 2y)\n\nSo y = 3x² →2y = 3(3x)²\n\nie y = 27x²/2
779655
This can be solved using Cardano's formula for the cubic equation.\n\nIt depends on the sign of 9-x^3/27\n\nIf 9-x^3/27 >0 there is one real solution and two complex conjugate solutions as follows:\n\nLet u =(-3 + √(9-x^3/27))^(1/3) and v =(-3 - √(9-x^3/27))^(1/3) \nThen y1 = u+v\ny2,3 = -(u+v)/2 ± (u-v)i√3/2\n\n\nif 9-x^3/27 <0 there are three distinct real solutions. (this is known as casus irreducibilis)\nif 9-x^3/27 = 0 there are three real solutions, two of which coincide.\n\nLet r = √(x^3/27)\nand cosφ = -3/√(x^3/27)\n\ny1 = 2r^(1/3)cos(φ/3)\ny2 = 2r^(1/3)cos(φ/3 + 120°)\ny3 = 2r^(1/3)cos(φ/3 + 240°)
106013
y = -2x + 8\n\nsolve it yourself