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How would you solve sin3x equals sinx?
To start, try breaking down sin3x using a double angle formula. Message me if you need more help!
How do you solve the limit 1 minus cos3x divided by sin3x as approaches 0?
Use l'Hospital's rule: If a fraction becomes 0/0 at the limit (which this one does), then the limit of the fraction is equal to the limit of (derivative of the numerator) / (derivative of the denominator) . In this case, that new fraction is sin(3x)/cos(3x) . That's just tan(3x), which goes quietly and nicely to zero as x ---> 0 . Can't say why l'Hospital's rule stuck with me all these years. But when it works, like on this one, you can't beat it.
Find the value of the limit x approaches 0 sin3x divided by tan6x?
That's 1/2 .(You have to use l'Hospital's rule.)
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
What is logic function-expression?
determine the constants a&b so that y=x(a sin3x + b cos 3x)wll satisfy identically the differential equation d2y/dx2+9y=6sin 3x
What is the integral for sin3x?
S sin3X dx =-1/3 cos3X
Integration of cos3 x?
int cos3x=sin3x/3+c
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 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
Solve sin3x plus cos3x equals 6?
If you want sin(3x) + cos(3x) = 6, then this is impossible. Sine and cosine will only return values between -1 and 1, so the expression sin(3x) + cos(3x) could only take values from -2 to 2, although even this is to great as sine and cosine of the same number will never both be 1 or -1. Similarly, if you want a solution to sin3x + cos3x = 6, then this is also impossible, because any power of a number between -1 and 1 will itself be between -1 and 1.
What is the solution sets for sin3x plus sin2x plus sinx equals 0?
The solitions are in degrees. You may convert them to degrees should you wish. x= 0,90,120,180,240,270,360
How would you solve sin3x equals sinx?
To start, try breaking down sin3x using a double angle formula. Message me if you need more help!
How do you integrate cot3x dx?
1/3ln(sin3x) + C
What does the sin3x equal?
sin(3x) = 3sin(x) - 4sin^3(x)
What is the integral of cot squared x?
ln(sinx) + 1/3ln(sin3x) + C
How do you solve the limit 1 minus cos3x divided by sin3x as approaches 0?
Use l'Hospital's rule: If a fraction becomes 0/0 at the limit (which this one does), then the limit of the fraction is equal to the limit of (derivative of the numerator) / (derivative of the denominator) . In this case, that new fraction is sin(3x)/cos(3x) . That's just tan(3x), which goes quietly and nicely to zero as x ---> 0 . Can't say why l'Hospital's rule stuck with me all these years. But when it works, like on this one, you can't beat it.
If sin3x equals 966 then what does x equal?
This has no solution. The sine of any number is always between 1 and -1.
