solving trigonometric equations
119380
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.
607936
(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
447896
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
414846
Have you learned trig identities yet?  I'd check to see if there was a trig identity you could substitute here.\n\nBTW, I am assuming you meant cos²x rather than cosx².\n\nTry this:  cos²x + sin²x = 1, so substitute 1-sin²x for cos²x:\n\n6cos²x + sinx - 4 = 0\n\n6(1-sin²x) + sinx - 4 = 0\n\n-sin²x + sinx + 2 = 0\n\n(multiply by -1 to make it a little easier)\n\nsin²x - sinx - 2 = 0\n\n(sinx-2)(sinx+1) = 0\n\nSo, sinx = 2 or sinx = -1\n\nSolving for x, this means x = sin^-1(2) or sin^-1(-1)
449074
If sin A = x and sinB = -x  then \nA = .... , -4π - B, -3π + B, -2π - B, -π + B,.-B, π + B, 2π - B, 3π + B, 4π - B, .... \n= nπ + (-1)^nB (n is any integer)\n\nSo arcsin(x) - arcsin(-x) = nπ + (-1)^nB - B\n= nπ + ((-1)^(n) -1)B\n\nNo this looks too clumsy ....new strategy!!\n\nsinA + sinB = 0\nSo 2 sin(½(A + B))cos(½(A - B)) = 0\n\nThus sin(½(A + B)) = 0 or cos(½(A - B)) = 0\nie ½(A + B) = 0, ±π, ±2π, ±3π, .... , ±nπ, \nie A + B = ±2nπ\nie arcsinx + arcsin (-x) = ±2nπ\n\nOr ½(A - B) =  ±π/2, ±3π/2, ±5π/2, ... , ±(2n - 1)π/2\nSo A - B = ±(2n - 1)π\nSo arcsinx - arcsin(-x) = ±(2n - 1)π
362293
tan x + tan 2x + tan 3x = 0\nor, tan x + tan 2x = -tan 3x\nor, sin x/cos x + sin 2x/cos 2x = -sin 3x/cos 3x\nor, (sin x *cos 2x + cos x * sin 2x )/cos x * cos 2x =  -sin 3x/cos 3x\nor, sin (2x + x )*cos 3x = - cos x * cos 2x * sin 3x\nor, sin 3x * cos 3x +  cos x * cos 2x * sin 3x = 0\nor, sin 3x (cos 3x + cos x * cos 2x)= 0\nor, sin 3x [cos ( 2x + x ) + cos x * cos 2x)= 0\nor, sin 3x [ cos x * cos 2x - sin x * sin 2x+ cos x * cos 2x)= 0\nor,  - sin 3x * sin x * sin 2x = 0\nor, sin x * sin 2x * sin 3x = 0\n\neither, sin 3x = 0\nie, 3x = nπ , n ε I\nie, x =  (nπ/3) , n ε I\n\nor, sin 2x = 0\nie, 2x =  nπ , n ε I\nie, x =  (nπ/2) , n ε I\n\nor, sin x = 0\nie,  nπ , n ε I\n\nBut putting x = nπ/2 , does not satisfy the equation\n(eg. Putting x = π/2, tan π/2 = undefined)\n& the solution set x = nπ is already included in the set x = nπ/3\nso the required solution is , x = nπ/3\n\nhence the general solutions of x are, x =  (nπ/3) n ε I ie. n = ....-3,-2,-1,0,1,2,3.......\n\n\nP.s. don't those dummies know about the significance of a G.S.\nI,m not telling it to you, you may make out the cause of such comment if you schroll down.
182452
cos^2 (2x) is equal to 1-sin^2(2x), remember, sin^2(x)+cos^2(x)=1
153932
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)
446137
Okay lets try this one ;\n\n2sin²  – 3sinx  + 1 →\n( sin x – 1)(2sin x – 1) = 0\nSin x – 1 = 0  →  sinx = 1 → x =∏ /2 \nX = 2∏ +∏ /2 , …., x = 2k∏ +∏/2 , ( k Є Z ) \n2sin x – 1 = 0  → sinx = 1/2 → x = ∏/6\nAs sin (∏ -∏/6) = sin ∏/6 = 1/2 \nSo ; \nX = 2∏ + ∏/6 , x = 4∏ + ∏/6 , …., x = 2k∏ + ∏/6 , ( k Є Z) \nX = 2∏ + 5∏/6 , x = 4∏ + 5∏/6 ,….., x = 2k∏ + 5∏/6 , ( k Є Z)\n\nGood Luck dear
500198
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
619857
A = 45 degrees or (pi / 4) radians\n\n1/sin^2A-1/cos^2A-1/tan^2A-1/cot^2A-1/sec^2A-1/csc^2A = -3\nIf i understand the way you wrote the problem above, it's 1/ each term, subtract each other right?\n1/ ((sinA)^2) - 1/ ((cosA)^2) - 1/ ((tanA)^2) - 1/ ((cotA)^2) - 1/ ((secA)^2) - 1/ ((cscA)^2) = -3\nI'm going to drop the A part, i assume you know that it will still be officially added in the final version\n\n[1/ (sin^2)] + [(-1)/ (cos^2)] + [-(cot^2)] + [-(tan^2)] + [-(cos^2)] +  [-(sin^2)] = -3\n[1/ (sin^2)] + [(-1)/ (cos^2)] + [-(cos^2) / (sin^2)] + [-(sin^2) / (cos^2)] + [-(cos^2) - (sin^2)] = -3\ncombine denominators\n[( 1 - (cos^2) ) / (sin^2)] + [( (-1) - (sin^2) ) / (cos^2)] + (-1) [(cos^2) + (sin^2)] = -3\n[(sin^2) / (sin^2)] + (-1)[( 1 + (sin^2) ) / (cos^2)] + (-1) [ 1 ] = -3\nTrig identity [(cos^2) + (sin^2)] = 1\n[ 1 ] + (-1)[( 1 + (sin^2) ) / (cos^2)] + [ -1 ] = -3\n(-1)[( 1 + (sin^2) ) / (cos^2)] = -3\n[( 1 + (sin^2) ) / (cos^2)] = +3\n1 + (sin^2) = 3 * (cos^2)\nTrig identity 1 = [(sin^2) + (cos^2)] =\n[(sin^2) + (cos^2)] + (sin^2) = 3 * (cos^2)\n2 * (sin^2) = 2 * (cos^2)\nsin^2 = cos^2\nsinA = cosA\nA= 45 degrees of (pi/4) radians
515224
Let a = A cos C\nb= A sin C\n\nsquare and add a^2+b^2 = A^2\n\nthen given expression\n= A cos C sin Bx + A sin C cos Bx = A sin(Bx+C)
476695
use formulas for cos(2x) and sin(2x) depending on cos(x) and sin (x) and then use the fact that\nsin(2x) = sqrt(1-cos(2x)^2).\nyou will get a quadratic equation that you solve with quadratic formula
137613
The final equation you mentioned should be actually:\n\nsin x = +/- sqrt(1 - cos^2 x)\n\nmeaning if you substitute the value of sin x as - sqrt(1 - cos^2 x) in the original equation which is sin^2 x + cos^2 x = 1, it should satisfy it, which it does.
342755
sin[x-(pi/2)]=kcosx\n-sin[(pi/2)-x]=kcosx\n-cosx=kcosx\nk=-1\ncos[x+{pi/2)]=ksinx\n=cos[(pi/2)+x]=ksinx\n=-sinx=ksinx\nk=-1
346551
tan(a+b)=sin(a+b)/cos(a+b)=\n(sinacosb+sinbcosa)/\n(cosacosb-sinasinb)\ndivide top and bottom by cosacosb\n(tana+tanb)/(1-tanatanb)
378931
L.H.S. (tanx - cotx)/(tanx + cotx) + 2cos^2x = (sinx/cosx - cosx/sinx)/(sinx/cosx + cosx/sinx) + 2cos^2x = (sin^2x - cos^2x)/(sin^2x + cos^2x) + 2cos^2x = (sin^2x - cos^2x)/1 + 2cos^2x = sin^2x - cos^2x+ 2cos^2x = sin^2x + cos^2x = 1 = R.H.S.