Sin2x=4900. How do I solve. I'm in calculus, but can't remember.?
463511
Use the double angle identities.
926243
sin^5(x) + cos³(x) = 1\n\nBy Quickmath (ref 1)\nx = 0, π/2 and 2π and a number of imaginery solutions \n\nYou can also solve it by the t formulae \nie if t = tanx/2 then sin x = 2t/(1 + t²) and cos x = (1 - t²)/(1 + t²)\n\nAfter some tedious algebra you get \n2t^2(t - 1)( polynomial of degree 7 in t) = o\n\nWhence t = 0, 0, 1 and some other irrational value.and 6 imaginery values\n\nSo x/2 = 0, π/4, π (in range 0 ≤ x/2 ≤ π)\nSo x = 0, π/2, 2π and the mystery value
983225
remember the rule nx^n-1\n2x^2-1\n \nSo in this case it is 2x
898718
Multiply both sides by x and by 5 (or just cross multiply):\n\nMultiply by x:\nx * sin(140) / 5 = sin(15)\n\nMultiply by 5:\nx * sin(140) = 5 * sin(15)\n\nDivide by sin(140):\nx = 5 * sin(15) / sin(140)\nx ≈ 2.01325478
956637
Sorry tried a lot, searched a lot & surfed a lot but found no solution, I think Pascal is right.\n\nP.s- Saket it is ∫e^sinx dx & not ∫(e^x)*sin x dx, if it was that, it would have been trivial.
1182723
sinx+cosecx=2\nsin+(1/sinx)=2\nfrom here try to cross multiply!
1102373
i wont answer.... log onto www.mathguru.com and discover solutions with explanations.
949593
Divide both sides by sin 20 to obtain n2.\n\nn2 = 1.0009 sin 30 / sin 20
765649
switch the variables.\n\nStart with\n\ny = sin(0.5x-3)\n\nand solve for x.\n\nto get rid of  the sin, we apply arcsin or sin^-1 (on calculators)\n\narcsin(y) = arcsin(sin(.5x - 3))\n\nthis gives\n\narcsin(y) = .5x - 3\n\nx = 2*[arcsin(y) + 3]
784821
Don't be faked out by the 2Q thing.\n\nLet x = 2Q\n\nThen the equation turns into sin x + cos x = 0\nor\nsin x = -cos x\n\nContemplate your unit circle for a while, and you should see that this is true when x = 135 degrees or x = 315 degrees.\n\nBut x = 2Q\nso Q = x/2\n\nHence, the answers are 67.5 degrees and 157.5 degrees.\n\nIf they want you to stick to the interval between 0 and 90 degrees, then the answer would of course be 67.5 degrees.
1185586
this has happened to me before on my 89. it's usually when the expression is too complicated--that is, it would take the calculator more memory than it has to solve it. so i'd recommend trying to first simplify the expression and then enter it. For instance, don't leave things in terms like sin(2) when you can just get an approximate number for that. maybe that's it?
397638
Use these identities:\nsin(2x) = 2*sin(x)*cos(x)\nsin(4x) = 4*cos^3(x)*sin(x) - 4*cos(x)*sin^3(x)\ncos(2x) = -1 + 2*cos^2(x)\ncos(4x) = 1 - 8*cos^2(x) + 8*cos^4(x)\ntan(3x) = (3*tan(x)-tan^3(x))/(1-3*tan^2(x))\n\nAfter you expand the left hand side of the equation with the above identities, you can factor out a 2*sin(x)*cos(x) from the numerator and a 2*cos^2(x) from the denominator, which will turn into a tan(x). Now divide the numerator and denominator of the fraction by cos^2(x). You should get the right hand side of the tan(3x) identity above.\n\nTry this and if you can't figure it out, I'll write it out. (I don't really want to since equations look like garbage on this site).\n\nCheck out this site for some usefull identities.
831435
The graph of sin x goes up to 1 when x = pi/2. When you graph y = sin (2x), it shrinks the x-values by half - so it will be 1 when x = pi/4.
814691
Did you get the question correct?\n\nStart from right-hand side:\n   cos^2x sinx - cos^4x\n= cos^2x [sinx - cos^2x]\n= cos^2x [sinx - (1 - sin^2x)]\n= cos^2x [sin^2x + sinx -1]\n\nYou can't transform (sin^2x + sinx - 1) into sin^3x.
748469
The first thing I would do is rewrite both a and s in terms of h.\n\nThen you use the following rules to find the derivative.\n\n1.  Use the quotient rule to find the derivative of  sin(a)/s^2.  If you don't know the quotient rule -- look it up and memorize it.\n\n2.  The derivative of the sin is the cosin -- use the chain rule to get cos(a)*da/dh\n\n3.  The derivative of the bottom is  2*s * ds/dh (again using the chain rule).\n\nOnce you get the derivative, solve for zero.\n\nGood luck.
880226
it is not what you specified\n\nIt cannot be solved without identities\n\nsin ^5 t = sin^4 t sin t\n= (1-cos^2 t)^2 sin t\n\nnow let cost = x\nso diffentiate to get - sint dt = dx\n\nso we need to integrate\n\n-(1-x^2)^2 dx\n\n= -dx + 2 x^2 dx - x^4 dx\n\nintegrating we get\n\n-x + 2x^3/3 - x^5/5 + C\n= - cos t + 2 cos^3 t/3 - cos^5 t /5 + C\n\nwhere C is constant of integration
772148
where did you get THIS problem?\nThe (real-valued) solutions are:\n\n3π/2 + 2πn and ≈ 2.945023353 + 2πn where n = 1,2,3,...\n(radians, of course)\n\n2sin2x - 2cosx - sinx = 1\n4sin(x)cos(x) - 2cos(x) - sin(x) = 1\n2cos(x)*[2sin(x) - 1] - sin(x) = 1\nlet y = sin(x) then cos(x) = √(1-sin^2(x)) = √(1-y^2)\nso,\n2√(1-y^2) *[2y - 1] - y = 1\nor\n2√(1-y^2) *[2y - 1] = 1 + y\nsquaring both sides,\n4(1 - y^2)(2y - 1)^2 = (1 + y)^2\n4(1 - y)(1 + y)(4y^2 - 4y + 1) = (1 + y)^2\nwe can divide both side by (1+y) noting that this shows that\ny = -1 is a solution, or sin(x) = -1, x = 3π/2 + 2πn ; n = 1,2,3,...\n\nthen we have\n4(1 - y)(4y^2 - 4y + 1) = (1 + y)\nexpanding,\n16y^3 - 32y^2 + 21y - 3 = 0\n\nThis is why I asked where you got this problem.  This cubic equation is not easily solved.  There are two complex conjugate solutions as well as y ≈ 0.195305854...(I wont show a derivation.  Use Mathematica or something similar. Plug it in-it works. You could also use Cardano's formula to find an exact value but I'm sure not doing that here and now.  Maybe if you paid me.)\n\nSo, sin(x) ≈ 0.195305854\nThen x ≈ 0.1965696 or π - 0.1965696 ≈ 2.945023353 \nBut the first one gets rejected since cos(x) must be negative.(This spurious solution comes from squaring the original equation)  Also, we can add even multiples of 2π to the solution to get another.  So,\nx ≈ 2.945023353 + 2πn where n = 1,2,3,...\n\nThese are the real solutions.  As I indicated, there are two complex conjugate solutions which will yield additional complex solutions for x. \nsin(x) ≈ 0.90234707 ± 0.3818408i\n\nYou can figure out x in these cases!!!