Cosx + cos2x = 0, the value of X

Cosx + cos2x = 0, the value of X

cosX+cos2X=0
cosx+2(cosx)^2-1=0
2(cosx)^2+cosx-1=0
(cosx+1)(2cosx-1)=0
Cosx = - 1 or cosx = 1 / 2
X = π + 2K π or - π + 2K π
Or x = π / 3 + 2K π or - π / 3 + 2K π

When x tends to 0, f (x) = e ^ X - (1 + ax) / (1 + BX) is the third order infinitesimal of X to find ab
It's the value of a and B, not ab

It can be seen from the meaning that LIM (x → 0) (f (x) / x ^ 3) = K (k is a non-zero constant)
Using the law of lobida, f '(x) / (x ^ 3)' = (e ^ x-a (1 + BX) - B (1 + ax)) / 3x ^ 2 = k
Let's use the lobida's law again, (e ^ x-2ab) / 6x = K
So LIM (x → 0) (e ^ x-2ab) = 0
ab=1/2
A and B are 1 / 2 ± I / 2 respectively (a + B = 1, ab = 1 / 2, we can know from the discriminant of root that it must be an imaginary number root)

When x tends to 0, e ^ (xcosx ^ 2) - e ^ X and x ^ n are infinitesimals of the same order, find the value of n!

e^(xcosx^2)-e^x/x^n=
e^(xcosx^2-x)-1 /x^n
=xcosx^2-x) /x^n=1
x(cosxx-1) 1/2x*(x^2)^2
n=5