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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
