Solving cos x = -x: What Does it Mean?

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This is only part of my question and I need the answer to continue, when does cos x = -x? on my calculator it is -.7390... isn't it supposed to be a nicer number ie pi over something?
 
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Nope. I'm afraid that's as good as it gets.
 
does y = x + cos x have any asymptotes? (horizontal or vertical) Vertical = none, the limit as x approaches infinity is infinity? And for negative infinity its negative infinity? is that possible since infinity is not in the domain of cos? or are there also no horizontal?
 
Infinities not numbers. You don't generally consider infinity in the domain of any function. So yes, there are no horizontal asymptotes either.
 
I believe we can show that this number is transcendental over \mathbb{Q}.
 
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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