How do you solve  2sin x = 4 sin^2 x?
927713
Let y = sin x.\n2y = 4y^2\n4y^2 - 2y = 0\ny(4y - 2) = 0\ny = 0 or 4y - 2 =0\ny = 0 or 1/2.\n\nSo sin(x) = 0 or 1/2. In degrees, using the range -90 to 90, this means that x is either 30 degrees or 90 degrees.
567245
simple x=r*cos(theta), y=r*sin(theta), r^2=x^2+y^2\nsubstitude, not hard
880856
Maybe you mean\n\n2sin^2(x) + 3sin(x) - 2 = 0\n(2sin(x) - 1)(sin(x) + 2) = 0\n\nBBQbaby agrees with my interpretation of the problem. There is no need to use the quadratic formula when the factoring is so easy.\n\nHer analysis of the answers makes sense.
949593
Divide both sides by sin 20 to obtain n2.\n\nn2 = 1.0009 sin 30 / sin 20
847419
This is a problem of trigonometric substitution.  The correct substitution is x = 2sin(t), as this will make the integrand become 2cos(t)dx.  Then, we have dx = d(2sin(t)) = 2cos(t)dt.  Thus, the integrand becomes 4(cos(t))^2.  The integral of this expression is 2t + sin(2t) + C.  We see from x = 2sin(t) that t = arcsin(x/2).  Hence, with substitution, we conclude that the desired integral is 2arcsin(x/2) + sin(2arcsin(x/2)) + C.\n\nObserve that\nsin(2arcsin(x/2)) =\n2sin(arcsin(x/2))cos(arcsin(x/2)) =\nx*root(1-sin(arcsin(x/2))^2) = \nx*root(1-(x^2)/4) =\n(x/2)*root(4-x^2)\n\nOur answer is therefore 2arcsin(x/2) + (x/2)root(4-x^2) + C.
129693
1) square both sides\n\n2) then make trig substitutions to get same 'variable'. Get sin^2[x] throughout equation. \n\ncos^2[x] + sin^2[x] = 1, so\n\n1 - cos^2[x] = sin^2[x].\n\n3) sin[x] cos[x] = ...\n\n4) Set equal to zero\n\n5) Solve.
1073595
(cosx)^2 - 2sinx = 2\ncosx^2 = 1 - sinx^2\n(1 - sinx^2) - 2sinx = 2\n1 - sinx^2 - 2sinx = 2\n-sinx^2 - 2sinx + 1 = 2\n-sinx^2 - 2sinx - 1 = 0\n-(sinx^2 + 2sinx + 1) = 0\n-(sinx + 1)(sinx + 1)\n\n-sinx - 1 = 0\n-sinx = 1\nsinx = -1\nx = sin^-1(-1)\nx = 270°\n\n--------------------------------------\n\n1.5^2 = 1.7^2 + 1.8^2 - 2(1.7)(1.8)cosA\n2.25 = 2.89 + 3.24 - 6.12cosA\n2.25 = 6.13 - 6.12cosA\n-6.12cosA = -3.88\ncosA = (388/612)\ncosA = (97/153)\nA = 50.6551127\n\n1.7^2 = 1.5^2 + 1.8^2 - 2(1.5)(1.8)cosB\n2.89 = 2.25 + 3.24 - 5.4cosB\n2.89 = 5.49 - 5.4cosB\n-2.6 = -5.4cosB\ncosB = (13/27)\nB = 61.2177953\n\n1.8^2 = 1.5^2 + 1.7^2 - 2(1.5)(1.7)cos(C)\n3.24 =  2.25 + 2.89 - 5.1cosC\n3.24 = 5.14 - 5.1cosC\n-1.9 = -5.1cosC\ncosC = (19/51)\nC = 68.1270919\n\nAs you can see, the angle opposite of side A, is the smallest angle.\n\nANS : about 51°\n\n----------------------------------------------\n\ncos(A) = 4/7\nx = 4\nr = 7\n\nr = sqrt(x^2 + y^2)\n7 = sqrt(4^2 + y^2)\n49 = 16 + y^2\ny^2 = 33\ny = sqrt(33)\n\nsin(A) = y/r\nsinA = (-sqrt(33))/7\n\ntanA = y/x\ntanA = (-sqrt(33))/4\n\ncscA = r/y\ncscA = -7/(sqrt(33))\ncscA = (-7sqrt(33))/33\n\nsecA = r/x\nsecA = 7/4\n\ncotA = x/y\ncotA = -4/(sqrt(33))\ncotA = (-4sqrt(33))/33
904134
sin3x = sin (x+ 2x) \n          = sinx.cos2x + cosx.sin2x\n          = sinx (2cos^2x -1) + cosx (2sinx.cosx)\n          = 2sinx.cos^2x - sinx + 2sinx.cos^2x\n          = sinx (2cos^2x - 1 + 2cos^2x)\n          = sinx (4cos^2x - 1)\n \n But, cos^2x = 1 - sin^2x\nso then,\n\n   sin3x = sinx [ 4 - 4sin^2x - 1]\n             = sinx (3 - 4sin^2x)\n\n Therefore,\n\n  sin3x/sinx =  sinx(3 - 4sin^2x)/sinx\n                   =  3 - 4sin^2x  --- answer.
1182723
sinx+cosecx=2\nsin+(1/sinx)=2\nfrom here try to cross multiply!
1059709
i think ur question meant sinx = cosecx\nsinx = 1/sinx\nsin^2x = 1\nsinx = 1\nsinx is 1 at 90degree\nthus  x = 90 degree
842198
I think you mean\n\n(sin^4 t - cos^4 t )/ sin 2t where t is for O\n= (sin ^2 t+ cos^2t)(sin^2 - cos ^ 2 t) as (a+b)(a-b) = a^2 - b^2\n= 1. (- cos 2t)/sin 2t  using sin^2t + cos^2 t = 1 and cos^2 t- sin^2 t = cos 2t\n= - cot 2t
732786
Using Trig identity 2sin(x)cos(x) = sin2x\n\n(2sin3x)(cos3x) = sin6x
129827
3.44(sinx)^2 -23.06=25sin2x\nsquare the 2 members u'll have\n11.8336(sinx)^4 -158.6528(sinx)^2+531.7636=\n25*4[(sinx)^2 * (1-(sinx)^2)]\ncoz sin2x=2sinx*cosx &(sin2x)^2= 4(sinx)^2(cosx)^2\nredice this equation :\n111.8336(sinx)^4 -258.6528(sinx)^2 +531.7636=0\nput t=(sinx)^2\n111.8336t^2 -258.6528t+531.7636=0\ndelta=7432.23321088\nt=0.77097759 we cannot take the other solution coz t<1\n=> sin x =+0.87805329 or sinx=-0.87805329\nx=61.408418 or x=-61.408418 or x= 118.59158196 or\nx= -118.59158196\nthere r other ways to solve it but hope this helps
913797
You can begin by converting sin x^2 to 1-cos2x and since they have the same denominator x^2 just split it in two integrals (1/x^2) - (cos2x/x^2).\nThe first one is easy to solve, do initegration by parts for the second integral.
791534
(2asinx*cosx)^2 + a^2(cos^2x - sin^2x)^2\n=4a^2 sin^2xCos^2x + a^2(cos^4x+sin^4x -2sin^2xCos^2x )\n=2a^2 sin^2xCos^2x + a^2cos^4x+a^2sin^4x\n=a^2(cos^2x + sin^2x)^2\n=a^2
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)
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)