X^3-2x-2cos(x) find local extrema

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Homework Statement

Find the local extrema of the function f(x)=(x^3-2x-2cos(x))

Homework Equations



derivatives, some algebra

The Attempt at a Solution



Well, the concept is simple.
Solve for the first derivative and set it equal to zero:

dy/dx=2sin(x)+3x^2-2=0 and

Next, solve for x to determine the "critical points".

My problem is in solving this seemingly simple equation algebraically.
I can simplify it to:
sin(x)=1-(3x^2)/2 (which doesn't help).

I have a feeling that it's not possible to solve algebraically, (but that I can still graph it).
Can anyone confirm my suspicion?

Thanks in advance!
 
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No, you can't solve that algebraically. Proceed with a graphical solution.
 
Thanks!
 

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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