Signal Analysis - Invertibility

FrogPad
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I don't understand why my book is saying that this system is not invertible.

y(t) = cos(x(t))

?

Wouldn't, arccos{ y(t) } = x(t) and therefore be an inverse system?
 
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First check on what your book says the definition of invertible is.

What about times T1 and T2, where T1 and T2 are different

x(T1) = 0, x(T2)= 2PI

y(T1) = 1 and y(T2)=1

Suppose

y(t)=1 what is the value of t?
 
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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