Solving Equations with Trigonometric Functions and Exponentials: Tips and Tricks

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Hi all,,

i hope you could help me in the two problem below:rolleyes:

1-2piCOS(2piX)=0

and

y=exp(x-1)-x
i want here x in terms of y

thanx for help:rolleyes:
 
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? It's not clear to me what you are asking. There is no "y" in the first equation so x cannot be written "as a function of y", it is a constant: cos(2\pix)= 1/(2\pi) so x= 1/(2\pi)cos^{-1}(1/(2\
pi)[/itex].

If the second equation is independent of the first, the x can be written as a y, but not using "elementary" functions. You should be able to write it in terms of "Lambert's W function" which is defined as the inverse function to f(x)= xex.
 
hi man and thank you for reply,,

its big problem for me and I am really want help:
the first equation is
f(x)=x-sin(2pix) and they want from me to find the absolute maximum and absolute minimum of f(x)?



the second equation is independent of the first
y=exp(x-1)-x

the question is
f(x)=exp(x-1)-x
find the inverse
so i want x=0.5y+1(for example)
i sure that you know the inverse, so i wait your answer

thank you for reply my dear:smile:
 
Broken_Mirage said:
the first equation is
f(x)=x-sin(2pix) and they want from me to find the absolute maximum and absolute minimum of f(x)?

There is no absolute max or min. \displaystyle\lim_{x\to\infty} x-\sin(2\pi x)=\infty and \displaystyle\lim_{x\to-\infty} x-\sin(2\pi x)=-\infty.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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