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To solve this equation, you must first put the equation in terms of cosx.
3cosx=2
cosx = 2/3
Next, you find the reference angle (α) by finding the cos inverse of 2/3.
α=cos-1(2/3) = 48.19 Degrees (approximately)
find the distance from from 48.19 and 90 you then add the difference to 270 giving you the two answers which are ... 48.19 and 311.81 (approximately)
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Find the derivative of y equals 3cosx?
y=3 cos(x) y' = -3 sin(x)
What is the derivative of cos-13 5cosx5 3cosx?
d(cos−13 + 5cosx5 + 3cosx)/dx = −1/2(1−32) − 25x4sinx5 − 3sinx
What is formula for cos3x?
looks like the exponents did not show up, in the first it should be 4 cosine cubed x - 3cosx and the sin 3x should be 3sinx - 4sine cubed x
How does one evaluate cosx plus 1 2cos2x-3cos2-3cosx-2 equals 0 looking for a walk through more than an answer... need to understand how this works to repeat the process on other problems?
The way you have written this makes it impossible to evaluate. what math operation is involved after cosx plus 1? is that added, subtracted, multiplied or divided by 2 cos2x-3cos2 etc.
What is the derivative of 3cosx?
The derivative of 3cos(x) is -3sin(x). This can be found using the chain rule, which states that the derivative of a composition of functions is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. In this case, the derivative of cos(x) is -sin(x), and when multiplied by the constant 3, we get -3sin(x) as the derivative of 3cos(x).
How do you express sin 3x as a polynom with sin x using the De Moivre formula?
According to de Moivre's formula, cos3x + isin3x = (cosx + isinx)3 = cos3x + 3cos2x*isinx + 3cosx*i2sin2x + i3sin3x Comparing the imaginary parts, isin3x = 3cos2x*isinx + i3sin3x so that sin3x = 3cos2x*sinx - sin3x = 3*(1-sin2x)sinx - sin3x = 3sinx - 4sin3x
What is the indefinite integral of 3sinx-5cosx?
8
What is the integral of cos cubed x?
cos^3(x) = cos^2(x) cos(x)= (1-sin^2(x))·cos(x)=cos(x) - sin^2(x)·cos(x)Integrating, the integral of cos(x) is sin(x) -----(1)To integrate sin^2(x) cos(x)·dx, let sin(x) =ucos(x)·dx = duThe integral becomes integral (u^2)·du = u^3/3 = (sin^3(x))/3 --(2)(1)-(2) gives the answer to your questionsin(x) - (sin3(x))/3 +C*Edit*When we are solving the integral of cos(x) - sin^2(x)·cos(x),& the two terms are split up, do not forget to take our signs into account, in this case;the positive or negative constants/coefficients within what we are taking the integral of.So here, ∫(cos(x) - sin^2(x)·cos(x))is split up into ∫cos(x) minus ∫sin^2(x)·cos(x)the 2nd integrand (function which is being integrated)-sin^2(x)·cos(x)is the same as(-1)·sin^2(x)·cos(x)So when taking the integral of this integrand, pull -1, a constant, out & place it in front of the integral symbol , ∫.So once we substitute u for sin(x) & integrate our integrand (original function) which we have now put into terms of u & du, and get (u^3)/3, as the integral of sin^2(x)·cos(x)) [again, (u^3)/3 is in terms of du, when u=sin(x)] we simply cannot ignore the fact that we are subtracting that integral from that of cos(x), no matter how easy it seems to just ignore our negatives.W.e are subtracting ∫(u^2)·du from whatever the ∫cos(x) is equal to, not adding them.sin(X) MINUS (sin^3(x))/3 + Cshould be the correct answer.ANOTHER WAY CAN BE SUBSTITUTING THE VALUE OF COS^3X FROM COS3X i.eINTEGRATE COS^3X....................................14COS^3X- 3COSX4COS^3X- 3COSX= COS3XCOS^3X=(COS3X+3COX)/3SUBSTITUTING THE VALUE OF COS^3X IN EQUATION 1INTEGRAL (COS3X+3COSX)/3AS 1/3 IS CONSTANT IT CAME OUT OF INTEGRALINTEGRAL OF COS3X=SINX/3INTEGRAL OF 3COSX= 3SINX..........................................................
Integral of (sin2x)(cos3x)?
Int[sin(2x)*cos(3x)]dx = Int[(2sinx*cosx)*(4cos^3x - 3cosx)]dx= Int[(8sinx*cos^4x - 6sinx*cos^2x)]dx Let cosx = u then du/dx = -sinx So, the integral is Int[-8*u^4 + 6*u^2]du = -8/5*u^5 + 2u^3 + c where c is a constant of integration = -8/5*cos^5x + 2cos^3x + c
What is the integral of sin x cubed?
= cos(x)-(cos3(x))/3 * * * * * Right numbers, wrong sign! Int(sin3x)dx = Int(sin2x*sinx)dx = Int[(1-cos2x)*sinx]dx = Int(sinx)dx + Int[-cos2x*sinx]dx Int(sinx)dx = -cosx . . . . . (I) Int[-cos2x*sinx]dx Let u = cosx, the du = -sinxdx so Int(u2)du = u3/3 = 1/3*cos3x . . . . (II) So Int(sin3x)dx = 1/3*cos3x - cosx + C Alternatively, using the multiple angle identities, you can show that sin3x = 1/4*[3sinx - sin3x] which gives Int(sin3x)dx = 1/4*{1/3*cos(3x) - 3cosx} + C
What is sin 3x in terms of sin x?
given the identity sin(x+y)=sinx cosy + siny cosxsin2x = 2 sinx cosx andsin(2(x)+x) = sin 2x cos x + sinx cos 2xusing the last two identities givessin3x= 2 sinx cosx cosx + sinx cos2xfactoring the sinx we havesin3x = sinx(2cosx cosx+cos2x)which satisfies the requirement.However, we can simplify further since cos 2x = cosx cosx - sinx sinx (a well known identity)sin3x = sinx (2cosx cosx +cosx cosx - sinx sinx)so sin3x= sinx(3cosx cosx - sinx sinx)or sin 3x = 3.cos²x.sinx - sin³x* * * * *Good, but not good enough. The answer was required in terms of sin, not a mixture of sinx and cosx. Easily recitified, though, since cos²x = 1 - sin²xTherefore sin3x = 3*(1-sin²x)*sinx - sin³x= 3sinx - 3sin³x - sin³x= 3sinx - 4sin³x
Is burnt wood a chemical change?
Yes, burning wood is a chemical change because it involves a chemical reaction that changes the composition of the wood. The heat and oxygen trigger the wood to undergo combustion, resulting in the release of new substances such as carbon dioxide, water vapor, and ash.
