solve the following equation :   4 cos 2*theta + 3 cos*theta = 0  for values between  0 and  180 degrees.?
831103
I would change cos 2 theta to 2 cos squared theta -1.\n\nYou then have a quadratic in cos theta.
567245
simple x=r*cos(theta), y=r*sin(theta), r^2=x^2+y^2\nsubstitude, not hard
287405
Use the identity cos 2x = 2 cos^2 x - 1\n\nThen you have 3(2cos^2 x -1) - 5cosx - 1 = 0\n\n6 cos^2 x - 3 - 5cosx - 1 = 0\n\n6 cos^2 x - 5 cosx - 4 = 0\n\n(3 cosx - 4)(2 cosx + 1) = 0\n\ncosx = 4/3 (impossible) or cosx = -1/2\n\nIn degrees, x = 120, 240, or if you need, 120+360n, 240+360n\nIn radians, x = 2pi/3 or 4pi/3 etc.\n\nHope that helps you see what to do.
276717
cos 45 degrees = (square root of 2)/2 \n               = about 0.70710678118654752440084436210485
1149867
We have x=rcos(θ), and y=rsin(θ).  \n\n\nTherefore x=4sin(θ)cos(θ) and y=4sin(θ)sin(θ).  \nAnd x^2+y^2= 16sin(θ)^2(cos(θ)^2+sin(θ)^2) =16sin(θ)=4y.  \nThus x^2+y^2-4y+4-4 =x^2+(y-2)^2-4=0 and x^2+(y-2)^2=4.\n\nAre you sure the second one is given by:\nρsin(θ)=2cos(θ) and not ρsin(φ)=cos(θ)?\n\nif it is ρsin(φ)=2cos(θ), then first notice that φ≠kπ for any integer k.\nThen we have ρ=2cos(θ)/sin(φ)\nNext we have x=ρcos(θ)sin(φ)=2cos(θ)^2, and y=ρsin(θ)sin(φ)=2cos(θ)sin(θ).\n\nThus x^2+y^2= 4cos(θ)^2(cos(θ)^2+sin(θ)^2) =2x,\nand x^2-2x+1-1+y^2= (x-1)^2-1+y^2=0,\nso (x-1)^2+y^2=1
256389
Let u=sin x, so\n4u^2 -4u +1 = 0 becomes\n(2u-1)^2=0\n2u-1=0\n2u=1\nu=1/2\nsin x=1/2\nx=30 degrees
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
857809
Hi, on your first one,\nsin^3 θ cos θ - sin θ cos^3 θ = -1/4 \n\nFactor out sin θ cos θ (sin^2 θ - cos^2 θ) = -1/4\n You will notice that if the part in parentheses were reversed it would be an identity for cos 2θ so multiply both sides by -1 to have this happen:\n\nsinθ cosθ (cos^2θ - sin^2θ) = 1/4\n\nYou will also notice if there were a 2 in front you'd have an identity for sin 2θ so multiply both sides by 2 to get\n\n2 sinθcosθ (cos^2θ - sin^2θ) = 1/2\n\nreplace with the identities: sin 2θ cos2θ = 1/2\n\nNow, if this had a 2 in front it would be an identiti for sin 2(2θ) so mult. both sides by 2 to get\n2 sin 2θ cos 2θ = 1\n\nsin 2(2θ) = 1\n\nsin 4θ = 1\n\nYou know that sin pi/2 = 1 so 4θ = pi/2 and θ = pi/8.  You can figure the additional answers as θ increases.\n \n\n If I get the second I'll edit this unless someone else already got it\n\nHere's what I think:\nsec 4θ + 2 sin 4θ = 0\n\n1/cos4θ + 2 sin 4θ = 0\n\nget common denominators\n\n1/cos 4θ + 2 sin4θ cos4θ/cos 4θ\n\nyou will see that the top of the second fraction is an identity for sin 2(4θ) so replace it\n\n1/cos4θ + sin 2(4θ)/cos4θ = 0\n\nThis is 1 + sin 8θ all over cos 4θ = 0\n\nIf a fraction = 0, its numerator = 0 so 1 + sin 8θ = 0\n\nsin 8θ = -1\n\nsin of 3pi/2 = -1 so 8θ = 3pi/2 and θ = 3pi/16\n\nagain there are infinitely many more answers unless you were given a specific domain for θ\n\nHope that helps
934584
2sin thete/2 cos theta/2=cos theta/2\n2sintheta/2cos theta/2-cos theta/2=0\ncos theta/2(2sintheta/2-1)=0\ncos theta/2=0=pi/2\ntheta/2=2npi-pi/2\ntheta=(4n-1)pi\n2sin theta/2-1=0\n2sin theta/2=1\nsin theta/2=1/2=pi/6\ntheta/2=npi+(-1)^n*(pi/6)\ntheta=2npi+(-1)^6*pi/3
571370
Assuming that the equation represents a circle where r is the radius take r on the other side ie. -sin theta=2/r , substitute value of r and use the natural sines table to get the proper angle.\n\neg. - 5 sin theta=2\n     therefore - sin theta=2/5=0.4\n    from natural sines table,in the row of 23 & column of 30' [angle is negative since Q. contains [-]sign]\n\nwe have 0.3987.Using mean diff. table, 5'=13\n     0.3987 + 0.0013 = 0.4001[approx 0.4]\n    therefore theta= -23 degrees and [30+5]' ['= minutes]
706305
R^2=A^2+B^2 -2(A*B)cos(180-acute angle)
310302
using @ for angle theta:\n\nx = r cos @, y = r sin @, use cos^2 @ + sin^2 @ = 1\n\nr^2 = 1/4*(1-5/4*sin^@)
821835
Problem 1.\nu = sin x\nu^2 + 3u + 2 = 0\nu = -3/2 +/- 1/2\nu = -1, -2\nsin x = -1  [because the sine is never -2]\nx = 270 degrees\n\nProblem 2.\nu = cos x\nu^2 + 2 u + 1 = 0\n(u+1)(u+1) = 0\nu = -1\ncos x = -1\nx = 180 degrees
791344
R = [ a^2 + b^2 -2*a*b* cos(theta), where  theta is the angle between a and b.\nIf a and b are fixed,\nR is max when  cos(theta) is min.   \nMin value of cos(thete) = -1  at  thete = 180 degree
247689
Use analytical trig (hint: sin function).  That should give you the answer you are looking for.
882948
Just make the following substitution:\nr=sqrt(x^2+y^2)\ntheta=arctan(y/x)\nand then simplify.  Bring everything over to one side and you get your answer.\n\nIf tan(theta)=y/x, then sec(theta)=sqrt(x^2+y^2)/x.\n\nThe problem is a little ugly, but it is doable and then at the end everything ends up canceling nicely.
422360
A triangle is 180 degrees... so 28 + 58 = 86. so now solve this: 180-86=....94!! Tah Dah!