Solve for x: 4sinxcosx + 2sinx - 2cosx - 1 = 0?
368536
2cosx(2sinx-1)+2sinx-1=0\n(2sinx-1)(2cox+1)=0\nsinx=1/2\ncosx=-1/2\nx= π/6 + 2πk, where k is a member of the set of integers\nx= 2π/3 + 2πk, where k is a member of the set of integers\nx= 5π/6 + 2πk, where k is a member of the set of integers\nx= 4π/3 + 2πk, where k is a member of the set of integers
225231
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
208014
Factorising gives\n\n(x-4)(x+1) = 0\n\nThe solutions are therefore x = +4 and -1
852832
x = [-b +/- sqrt(b^2 - 4ac)] / (2a)\nf(x): a = 1, b = 0, c = -4\ng(x): a = 1, b = -1, c = -12
820839
x+4+x= -20\nFirst collect like terms.\n2x+4= -20\nGet rid of the 4. Subtract 4 to both sides.\n2x+4-4= -20-4\n2x= -24\nIsolate the x. Divide both sides by 2.\n2x/2= -24/2\nx= -12
413913
x=0,2,3,4 and y=1,-1,0,3
324244
sinx + xcosx  (product rule)
412551
-4x+16=-1-4x\n-4x+4x=-16-1\n0=-17\n\nthe answer is [a]0
971568
What's the point of asking the same question twice?\n\nIf the issue is that you're only allowed to work on one side of the equation or the other, then you can pick one side, follow down to the bottom, and do everything in reverse to match the other side.\n\nRHS:\n\n1 + tan²x / tanx = [1 + (sec²x - 1)] / tanx = sec²x / tanx\n= 1/cos²x / (sinx / cosx) = 1/cos²x * cosx/sinx\n= 1/(sinx cosx) = cscx secx\n\nLHS:\n\n[cscx + secx]/[sinx + cosx] = [1/sinx + 1/cosx]/[sinx + cosx]\n= [(cosx + sinx)/(sinx cosx)] * 1/[sinx + cosx]\n= 1/(sinx cosx) = cscx secx
859883
cosx(1-2sin^2x)=0\ncosx(cos2x)=0\ncosx=0\nx=pi/2,3pi/2\ncos2x=0\n2x=pi/2 or 3pi/2\nso x=pi/4 or 3pi/4
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
825795
OK,as the first one is already done,I am doing the rest for you\n4(2x-6)=x+2 =>8x-24=x+2  =>8x-x=2+24  =>7x=26  =>x=26/3=8 2/3  (3) 2(x+5)+1=3(x-1)+4   =>2x+10+1=3x-3+4 =>2x-3x=-3+4-10-1  => -x=-4  =>x=1  (5)x+6=4(4x-1)   =>x+6=16x-4  =>x-16x=-4-6  => -15x=-10  =>x=-10/-15=2/3 [after reducing]
929299
2sinx+3=4\n2sinx=1\nsinx=1/2\nx = pi/6
901670
(sin x +cos x)^2 =1    [0=<x=<360]\nsin^2x + 2sinx * cosx +cos^2x =1 or 2sinxcosx=0      Because sin^2x+cos2^x =1 \n2sinxcosx is zero when x =0, 90, 180, or 270 degrees\n\n1-sinx = cosx     [0=<x=<360]\nsinx + cosx = 1\nThis occurs when x= 0, 360
901088
4sinxcosx =1 (assumed for 0 ≤ x ≤ 2π)\nSo 2 sin2x = 1 (since sin2x = 2sinxcosx)\nSo sin2x = ½ (0 ≤ 2x ≤ 4π)\n2x = π/6, 5π/6, 13π/6, 17π/6\nThus x = π/12, 5π/12, 13π/12, 17π/12\n\nsin4x= -2sin2x (assumed for 0 ≤ x ≤ 2π)\nSo 2sin2xcos2x + 2sin2x = 0\nie 2sin2x(cos2x + 1) = 0\nso sin 2x = 0 or cos 2x = -1/2 (0 ≤ 2x ≤ 4π)\n2x = 0, π, 2π, 3π, 4π or 2x = 2π/3, 4π/3, 8π/3, 10π/3\nwhence x = 0, π/3, π/2, 2π/3, π, 4π/3, 3π/2, 5π/3, 2π
809968
(tan x- cot x)/( tan x + cot x)+ 2cos^2 x=1\n\nSenx / cos x  - cosx/senx\n---------------------------------  + 2cos^2x\nSenx / cos x  + cosx/senx\n\n(sen^2 x - cos^2x) / senx cosx\n--------------------------------------  + 2cos^2 x\n(sen^2 x + cos^2x) / senx cosx\n\nThe fraction reduces on the above:\n\nsen^2 x - cos^2 x                       + 2 cos^2 x\n\nsen^2x + cos^2x = 1\n\nproved!
757614
4 (x^2 + 3X -4)\n4(x+4)(X-1)