Is Cos(pi(2x+1)) the Key to Simplifying the Search for x Values?

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sec(pi)=2/2x+1
-1=2/2x+1
-2x-1=2
-2x=3
x=-3/2

is this correct?
 
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is the original supposed to be sec(pi)(2x+1) = 2?
 
sec(pi(2x+1)=2

the original
 
That doesn't make any sense. You have two "(" and only one ")".
The whole question is whether you mean sec(pi(2x+1))= 2 or sec(pi(2x)+ 1= 2 and you still haven't answered that.
 
sec(pi(2x+1))= 2 is what i mean.
 
Karma said:
sec(pi(2x+1))= 2 is what i mean.

So cos(pi(2x+1)) = ? Does that suggest an easier way to search for the values of x?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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