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How do you integrate sinx divide sinx plus cosx?
1/2(x-ln(sin(x)+cos(x)))
How do you solve sin2x plus sinX equals 0?
Sin(2x) + Sin(x) = 0 2Sin(x)Cos(x) + Sin(x) = 0 Factor 'Sin*(x)' Sin(x) (2Cos(x) + 1 = 0 Hence Sin(x) = 0 x = 0 , 180(pi) , 360(2pi) et.seq., & 2Cos(x) = -1 Cos(x) = -1/2 = -0.5 x = 120, 150, 480, et.seq.,
How do you find the general solutions of csc x equals -2?
Cosec x = -2 => sin x = -0.5 The primary solution is x = -pi/6 radians. Therefore the solutions are: 2n*pi - pi/6 and (2n+1)*pi + pi/6 for all integer n.
Using the remainder formula for the Taylor polynomial approximation estimate the error in your approximation to sin n?
If you use n terms from the Taylor expansion, the absolute value of the error is less than [|x|^(2n+1)]/(2n+1)!If you use n terms from the Taylor expansion, the absolute value of the error is less than [|x|^(2n+1)]/(2n+1)!If you use n terms from the Taylor expansion, the absolute value of the error is less than [|x|^(2n+1)]/(2n+1)!If you use n terms from the Taylor expansion, the absolute value of the error is less than [|x|^(2n+1)]/(2n+1)!
What would be the integral of x where x has power 2?
If x has the power 2 then you want the integral of x2, I think. When you integrate this you get : x3/3 , plus a constant.
