Calculus- Finding x values of a function.?

Question

Consider
f(x) = cos^2x + 2cosx over one complete period beginning with x=0.

Find all values of x in this period at which f(x)= 0.

This is the first part of this problem and I have no idea how to solve for x with this. help please :)

Answer 1

let z = cos (x)

then the problem becomes
f(x) = z^2 +2z =0

the solution to this is z = 0 or z = -2

Since cos(x) can never be -2 we use the z = 0 solution

which means we have to solve cos(x) = 0
this has solutions at x = pi/2 and x = 3*pi/2

SO these are the two solution for the first period

Answer 2

Period = [ 0, 2pi]

f(x) = cos^2x + 2 cosx

for f(x) = 0

cos^2x = 2 cosx

dividing by cos x (excluding x = pi/2, which makes cos x = 0)

cos x = 2,

so there is no real solution (absolute value of cos x function = 1)

Answer 3

Drawing it would help some, actual, you recognize that your function is of degree 3, so which you have 3-a million = 2 'inflection' factors. those will take place whilst your first derivative is 0. To get an concept of what it rather is going to look like, you may look on the applications habit at +/- infinity. x - > + infinity , g --> + infinity ( the dominant term may be the x^3 one) x - > - infinity, g - > - infinity those factors are given by using, g ' (x) = 3 x^2 - 8x - sixteen= 0 remedy to locate that the roots are at, x = 4 x = - 4/3 So in case you recognize you may desire to locate a rel min/max evaluate g(x) at those factors g(4) = - sixty 9 g(-4/3) = 185/27 so -4/3 is the rel max and -sixty 9 the rel min merely evaluating their values. you additionally can examine this utilising the 2nd derivative. you will see that at -4/3 the 2nd derivative is unfavourable and at 4 the 2nd derivative is helpful wisely. you could convince your self that case in point, if at a interior sight extrema, the 2nd derivative is helpful.(first derivative is 0) this means the 1st derivative is increasing from 0. meaning the function could be increasing in the useful direction and subsequently a minimum!

Answer 4

If f(x)= cos^2x + 2 cosx then
f(x) = cosx( cos2x +2)
So f(x) = 0 when cosx = 0 or cos2x = -2