When the derivative is set to zero, the solution of the resulting equation gives the stationary points of f(x).
f'(x)=3x2-2+2sin(x)=0 at stationary points.
To obtain an approximate solution we can substitute sin(x)=x (an approximation). So we have 3x2-2+2x=0. Using the quadratic formula, x=(-2±√(4+24))/6=0.55 and -1.22 approx. Later we'll see that these rough solutions are quite close to the actual solutions.
f'(x)=3x2-2+2sin(x)=0 can be solved using Newton's Method, for which we'll need the second derivative:
f"(x)=6x+2cos(x).
xn+1=xn-f'(xn)/f"(xn) is an iterative process which gives iterations of x which (hopefully) converge to a single value.
To get the process started we need to choose a value for x0. We could choose x0=0, then:
x1=0-f'(0)/f"(0)=-(-2)/2=1 (radian), x2=1-(1+2sin(1))/(6+2cos(1))=0.6210...,
x3=0.5611..., x4=0.5593..., x5=0.5593721..., x6=0.5593721708=x7. So we have a solution for x=0.5594 approx.
f(0.5594)=-2.6389 approx. f"(0.5594)=5.05 which is positive so (0.5594,-2.6389) is a minimum.
There is likely to be another stationary point, so how can we find it?
Let's consider f(x)=x3-2x-2cos(x). We know that cosine and sine lie between -1 and 1, so replace cos(x) or sin(x) with -1 and 1:
f(x)=x3-2x-2 or x3-2x+2, f'(x)=3x2-2-2 or 3x2-2+2.
f'(x)=0 when x2=4/3 so x=±√(4/3) or x=0 as a rough approximation. We already used x0=0 (and x1=1) to find the solution above so let's try using x0=-1 in Newton's Method:
x1=-1.1388..., x2=-1.1264..., x3=-1.12629..., x4=-1.126299941=x5. So we have convergence to a different value x=-1.1263 approx.
f(-1.1263)=-0.03618, f"(-1.1263)<0 so we have a maximum at (-1.1263,-0.03618).
Compare the two solutions for x we obtained earlier: x=0.55 and -1.22,
In a cubic equation which f(x) resembles we would expect two stationary points so we appaear to have found them!