Trigonometry question.  How do I get the second solution for secx+4=0?
952703
The calculator's answer for inverse cosine will only be in quadrant I or II (yours is in Q II). Cosine is also negative in quadrant III.  The angle in Q III whose cosine is -1/4 (to the nearest hundredth of a radian) is 4.46.  (2pi - 1.82)\nSo you also have x = 4.46 + k (2pi)
310146
i think u can do this:\n               \n            (secA+tanA)2\n   =sec2A-2tanA.secA+tan2A\n  now,   put the value of secA that can b derived from the question\n             secA+tanA=2\n       or,secA=2-tanA\n                =sec2A-2tanA.2-tanA+tan2A\n           =sec2A-4tanA-2tan2A+tan2A\n             =sec2A-4tanA-tan2A\n              =tan2A-1-4tanA-tan2A\n             =-(1+4tanA)\n\nthis may b the ans(in the above solution in some places 2 means square,i hope u'll understand it)
1155939
It's something like:\n\n2x + 4 = 2(x +2)\n\nWhen you distribute the right hand side and then put the x's together and the constants together, you end up with 0=0.\n\nThat means any value of x would make this a true equation. There is no specific solution to this equation.\n\nLater, when you get into trigonometry, identities become very important and aren't as simply and obvious as this example. For example you learn that tan(x)=sin(x)/cos(x) and that is always true. (Of course, you don't know what that means yet.) These identities mean we can always substitute what's on one side of the equation for what's on the other side of the equation without changing anything.
876211
2 cos^2 x + sin^2 x = 1\n\ncos^2 x + cos^2 x + sin^2 x = 1\n\ncos^2 x + (cos^2 x + sin^2 x) = 1\n\ncos^2 x + 1 = 1\n\ncos^2 x = 0\n\ncos x = 0\n\nx = pi/2 or 3 pi/2
929231
(a^2-b^2)=(a-b)(a+b)\nHence,\nsec^4y-tan^4y = (sec^2y-tan^2y)(sec^2y+tan^2y)\n\nSubstitute that back into the initial equation...\n\n(sec^2y-tan^2y)(sec^2y+tan^2y)\n------------------------------------------\n           (sec^2y+tan^2y)\n\nand cancel common terms\n\n= (sec^2y-tan^2y) \n= 1 (a standard trig identity)\n\nhope that solves it.
772148
where did you get THIS problem?\nThe (real-valued) solutions are:\n\n3π/2 + 2πn and ≈ 2.945023353 + 2πn where n = 1,2,3,...\n(radians, of course)\n\n2sin2x - 2cosx - sinx = 1\n4sin(x)cos(x) - 2cos(x) - sin(x) = 1\n2cos(x)*[2sin(x) - 1] - sin(x) = 1\nlet y = sin(x) then cos(x) = √(1-sin^2(x)) = √(1-y^2)\nso,\n2√(1-y^2) *[2y - 1] - y = 1\nor\n2√(1-y^2) *[2y - 1] = 1 + y\nsquaring both sides,\n4(1 - y^2)(2y - 1)^2 = (1 + y)^2\n4(1 - y)(1 + y)(4y^2 - 4y + 1) = (1 + y)^2\nwe can divide both side by (1+y) noting that this shows that\ny = -1 is a solution, or sin(x) = -1, x = 3π/2 + 2πn ; n = 1,2,3,...\n\nthen we have\n4(1 - y)(4y^2 - 4y + 1) = (1 + y)\nexpanding,\n16y^3 - 32y^2 + 21y - 3 = 0\n\nThis is why I asked where you got this problem.  This cubic equation is not easily solved.  There are two complex conjugate solutions as well as y ≈ 0.195305854...(I wont show a derivation.  Use Mathematica or something similar. Plug it in-it works. You could also use Cardano's formula to find an exact value but I'm sure not doing that here and now.  Maybe if you paid me.)\n\nSo, sin(x) ≈ 0.195305854\nThen x ≈ 0.1965696 or π - 0.1965696 ≈ 2.945023353 \nBut the first one gets rejected since cos(x) must be negative.(This spurious solution comes from squaring the original equation)  Also, we can add even multiples of 2π to the solution to get another.  So,\nx ≈ 2.945023353 + 2πn where n = 1,2,3,...\n\nThese are the real solutions.  As I indicated, there are two complex conjugate solutions which will yield additional complex solutions for x. \nsin(x) ≈ 0.90234707 ± 0.3818408i\n\nYou can figure out x in these cases!!!
388285
equate the two equations: \nsin 2x = pi/4 \n2x = inverse sine pi/4 + k2pi or (pi - inverse sine pi/4) + k2pi\n2x = 0.90 + k2pi or 2.24 + k2pi\nx = 0.45 + kpi or 1.12 + kpi\nx = 0.45, 1.12, 3.59, 4.26 ( k = 0,1)
554578
Use the substitution x = tan(y) => x^2 +1 = sec^2(y) and\ndx = sec^2(y) dy.  After making the substitutions, you get \ntan^2(y)sec^2(y) dy in the numerator and\n[sec^2(y)]^3 = sec^6(y) in the denominator.\nCancel the sec^2(y) term and you are left with tan^2(y)dy in the numerator and sec^4(y) in the denominator.  Now, change the numerator to sec^2(y) - 1.  After that, break it up into two fractions, the first being sec^2(y)/sec^4(y) = 1/sec^2(y) = cos^2(y) and the second being 1/sec^4(y) = cos^4(y).\n\nSo, all you have to integrate is cos^2(y) - cos^4(y).  The easiest way is to factor out a cos^2(y) from both and rewriting it as cos^2(y)[1-cos^2(y)] = cos^2(y)sin^2(y) = 1/4[4cos^2(y)sin^2(y)] = 1/4[2sinycosy]^2 = 1/4[sin^2(y)].  Now, since sin^2(y) does not have an antiderivative, you must use the fact that sin^2(y) = 1/2{1-cos2y}.  Now, our integral becomes (1/4)(1/2)[1-cos4y]dy and we can integrate this to be 1/8 y - 1/8 [sin4y/4} = 1/8 y - 1/32 sin4y. Now, let's rewrite sin4y so that we can convert back to our original variable, x.  Using the double angle formula for sin, we have sin4y = 4sinycos^3(y) - 4sin^3(y)cos(y).  Now, we can rewrite this in terms of x as:\n\n1/8 y - 1/32{4[sinycos^3(y) - sin^3(y)cosy]} =\n1/8 arctanx -1/8[x/(x^2 + 1)^2 - x^3/[x^2 + 1]^2 + C.\nLet me know if you need further help.
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
943226
I'm only going to calculate the principal values.\n\nLet x = cos^[-1](sqrt(2) / 2).\nTherefore, cos(x) = sqrt(2) / 2.\nSo, x = pi / 4 radians.\n\nLet y = sin[-1](-1).\nTherefore, sin(y) = -1.\nSo, y = - pi / 2 radians.\n\nYour expression is then : sec(x + y)\n= sec(pi / 4 - pi / 2)\n= sec(- pi / 4)\n= 1 / cos(- pi / 4)\n= 1 / [sqrt(2) / 2]\n= 2 / sqrt(2)\n= sqrt(2)
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.
768749
A'ighty, first write sec(x) = 1/cos(x).  Now take the derivative using the Quotient Rule:\n\nd/dx (1/cos(x)) = (cos(x)*d/dx(1) - 1*d/dx(cos(x))) / cos^2(x)\n\n                        = (0 - (-sin(x))) / cos^2(x)\n\n                        = sin(x) / cos^2(x)\n\n                        = (1/cos(x)) * (sin(x) / cos(x))\n\n                        = sec(x)*tan(x).\n\nAnd there you go.  ^_^
64064
put, for example: t=cos x, then sin^2(x)= 1-cos^2(x)= 1-t^2\n\nThus you get from the original equation, a quadratic equation:\n\nt-2(1-t^2)+1=0 or t-2+2t^2+1=0, or\n\n2t^2+t-1 = 0.\n\nAfter factorization of the left side \n\n(2t-1)(t+1)=0\n\nthen 2t-1=0 gives one solution for t;\n\nt+1= 0 gives another solution.\n\nSince this is quadratic equation, you have received all the possible solutions.\n\nT_1 = 1/2 ;    T_2 = -1\n\nBut we have assumed earlier, to get to that equation, that\n\nT = cos (x)\n\nSo now we have to consider both options for t;\n\n1)cos x = 1/2; 2)cos x = -1\n\nFrom it goes something like that, check those calculations for yourself of course:\n\nSolution of the first is x = PI/2+ PIK, K= 0, 1..\n\nSolution of the second is x= PI+ PIK, K=0,1...\n\nThat is how you turn a trig problem into an algebraic one.\n\nHope that helps.
470919
I think you need to find A, if it is so then:\ntan A + sec A=2     square both sides\ntan^2 A + 2 tan A sec A + sec^2 A=4  since, sec^2 A= 1+tan^2 A  then:\n2 tan^2 A + 2 tan A sec A=3      since tan A=sin A/ cos A & sec A=1/cos A   then:\n2 sin^2 A/ cos^2 A + 2 sin A/ cos^2 A=3\n sin^2 A/ cos^2 A + sin A/ cos^2 A=3/2\n sin^2 A + sin A= 3/2* cos^2 A = 3/2* (1-sin^2 A)    since sin^2 A + cos^2 A =1\n5/2* sin^2 A + sin A -3/2= 0\n5 sin^2 A + 2 sin A – 3 = 0\n(5 sin A -3) (sin A +1)= 0 then,\neither sin A= 3/5 → A=36.87o \nor sin A=-1 → A=-90o (to be neglected)\ncheck: tan 36.87o=0.75 & cos 36.87o = 0.8 → sec 36.87o =1.25 then,\ntan 36.87o  + sec 36.87o =2  Q.E.D
796906
too complicated to type it up, sorry. look up what other ways can sec^2x  be written, and start substituing those in. good luck
510441
Divide both sides by cosX. We get 8sinX = 4. Now divide both sides by 8, then reduce. We get sinX = 1/2.\n\nThere are two solutions when looking at the sine curve:\n\nx = pi/6\nx = (5*pi)/6\n\nits very easy, make an honest effort to understand trig, and when you do the payoffs are exponential!
699547
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