Trigonometry.  Sin2x = 4900
209620
Use the double angle identities.
196518
Change cscx to 1/sinx and sin(-x) = -sinx\ncscx(cscx + sin-x)\n= (1/sinx)(1/sinx - sinx)\nGet a common denominator\n= (1/sinx)([1-(sinx)^2]/sinx)\n1-(sinx)^2 = (cosx)^2\nSo\n(1/sinx)([1-(sinx)^2]/sinx)\n= (1/sinx)((cosx)^2/sinx)\n= (cosx)^2/(sinx)^2\n= (cosx/sinx)^2\n= (cotx)^2
91713
sinx = a     with    a Є [-1;1]\n\n2sin^2 x - 5sinx + 2 = 0 \n<=> 2a^2 - 5a + 2 = 0\nWe have 2 roots: a = 2 (reject) & a= 1/2 because a Є [-1;1]\n => sinx = 1/2\n<=> x = ∏/3 + k2∏ and x = 2∏/3 + t2∏ with k,t Є Z
493657
6000 = 2500 * sin ((90/13)*x)+4500__________ -4500\n                        1500 = 2500 * sin ((90/13)*x) __________/2500\n                            .6  =  sin ((90/13)*x) _______________arcsin\n                arcsin (.6) = (90/13) * x ___________________ *(13/90)\n  (13* arcsin (.6))/90 = x\n\nx =0.0929501602
442677
I'll give you a hint and let you try to figure out the rest yourself (unless someone else tells you the exact answer). =)\n\nUse your trig identities:\n\ntan A = sinA/cosA\n\ncot A = 1/tanA or cot A = cosA/sinA
645632
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
409042
The x component is 12 cos 60 and the y componet is 12 sin 60.\n\nThat is x component is 6N and y component is 10.4 N
543602
sin^2(x)=1/2sin(2x)\nsin^2(x)*2sin(2x)=0\nsin^2x=0\nsinx=0\nx=npi\nsin(2x)=0\n2x=npi\nx=npi/2
335944
sinx/(1+cosx)\n= sinx(1-cosx)/[(1+cosx)(1-cosx_]\n= sinx(1-cosx)/(1-cos^2x)\n= sinx(1-cosx)/sin^2x\n=(1-cosx)/sinx\n= 1/sinx - cosx/sinx\n\n9+8cosx/sinx + 8sinx/1+cosx\n= 9+8[cosx/sinx + sinx/1+cosx]\n= 9 + 8[cosx/sinx + 1/sinx - cosx/sinx]\n= 9 + 8/sinx
390092
Finding max's and mins is done thru the first derivitive and setting it to 0, so\n\nf(x)= sinx - cosx\nf1(x)= cosx + sinx\n0=cosx + sinx\n-cosx = sin x\nthis occurs when x= (3pi)/4\nmax is root 2, as was stated.\n\nThe answer that used to be directly above is wrong.
512385
sinx+cosx*(cosx/sinx)=cscx\n(sin^2x+cos^2x)/sinx=cscx\nlet us check the problem\nsubstituting th values for x=30\nLHS=(1/2)+(rt3/2)(rt3)\n=1/2+3/2=2\ncsc30=2\nthough this is a solution since sin^2x+cos^2x is an identity the proble will have infinite solutions\nfor example if x=90*\nLHS=1\nand RHS=1\nfor x=60*\nLHS=rt3/2+(1/2)(1/rt3)\n=3+1/2rt3=2/rt3\ncsc60*=2/rt3
353956
(sinX + 1) = 0---> sinX = -1--->x = 3pi/2 + (2pi)n, radians, or x = 270 + 360n degrees, where n is any whole number including 0. \n\nThe first step was simple algebra.  In the second step, I just remembered that the sine is how far up on the unit circle that angle points, so -1 means it points straight down, which is 3/4 of a circle.  Adding a whole circle puts it right back where it started from.\n\ncosX + 1/2 = 0---> cosX = -1/2 ---> x = 2pi/3 or 4pi/3 + (2pi)n radians, 120 or 240 + 360n degrees.  \n\nAgain, the first step was simple algebra.  As for the second, my trig teacher made us memorize the sines/cosines of 30, 45, and 60 degrees, so I just know that the cosine of 60 degrees is 1/2.  Go around to the other side of the unit circle and the cosine of 120 degrees is -1/2.  Go down to the bottom of the unit circle, and the cosine of 240 degrees is also -1/2.
510098
sin(θ) ≈ θ\nSo cos (θ) ≈ 1  because cos (0) = 1\nMore accurately cos (θ) ≈ 1 - θ²/2
182452
cos^2 (2x) is equal to 1-sin^2(2x), remember, sin^2(x)+cos^2(x)=1
89533
sin2x=1/2\n=>2X=30\n=>X=15+360.K\nK={0,1,2...}
288281
sin(48) = 4/c\nc = 4/sin(48)\nside a can then be figured out using the pythagorean theorem.
577643
Use the Law of Sines.\n\nsin A/a = sin B/b\n(sin 36)/24 = sin B/47\nsin B = (47 sin 36)/24\nsin B = 1.151079452\nAt this point you have a problem--it is impossible for the sine of an angle to have a value greater than 1.\n\nSo either you copied the problem wrong, or you are being asked to recognize that a triangle cannot be formed with these dimensions.