We have: #f(x) = 5 cos^(2) (x) - 10 sin(x)#; #x in [0, 2 pi]#
In order to determine the points of inflection, we need to evaluate the second derivative of the function:
#=> f'(x) = (d) / (dx) (5 cos^(2)(x)) - (d) / (dx) (10 sin(x))#
#=> f'(x) = (5 cdot - sin(x) cdot 2 cos(x)) - (10 cdot cos(x))#
#=> f'(x) = - 10 sin(x) cos(x) - 10 cos(x)#
#=> f''(x) = (d) / (dx) (- 10 sin(x) cos(x) - 10 cos(x))#
#=> f''(x) = (- 10 cdot (sin(x) cdot - sin(x) + cos(x) cdot cos(x))) - (10 cdot - sin(x))#
#=> f''(x) = - 10 (- sin^(2)(x) + cos^(2)(x)) + 10 sin(x)#
#=> f''(x) = - 10 (cos^(2)(x) - sin^(2)(x)) + 10 sin(x)#
The points of inflection occur at those points where the second derivative is equal to zero:
#=> f''(x) = 0#
#=> - 10 (cos^(2)(x) - sin^(2)(x)) + 10 sin(x) = 0#
One of the Pythagorean identities is #cos^(2)(x) + sin^(2)(x) = 1#.
We can rearrange this to get:
#=> cos^(2)(x) = 1 - sin^(2)(x)#
Let's apply this rearranged identity to get:
#=> - 10 ((1 - sin^(2)(x)) - sin^(2)(x)) + 10 sin(x) = 0#
#=> - 10 (1 - 2 sin^(2)(x)) + 10 sin(x) = 0#
#=> - 10 + 20 sin^(2)(x) + 10 sin(x) = 0#
Let's rearrange this to form a quadratic equation:
#=> 20 sin^(2)(x) + 10 sin(x) - 10 = 0#
#=> 10 (2 sin^(2) (x) + sin(x) - 1) = 0#
#=>2 sin^(2) (x) + sin(x) - 1 = 0#
Then, let's factorise using the middle-term break:
#=> 2 sin^(2)(x) + 2 sin(x) - sin(x) - 1 = 0#
#=> 2 sin(x) (sin(x) + 1) - 1 (sin(x) + 1) = 0#
#=> (sin(x) + 1) (2 sin(x) - 1) = 0#
We now have a product that is equal to zero, so either one of the multiples must be equal to zero:
#=> sin(x) + 1 = 0#
#=> sin(x) = - 1#
#=> x = arcsin(- 1)#
#=> x = (3 pi) / (2) + 2 pi n#
or
#=> 2 sin(x) - 1 = 0#
#=> 2 sin(x) = 1#
#=> sin(x) = (1) / (2)#
#=> x = arcsin((1) / (2))#
#=> x = (pi) / (6) + 2 pi n, (5 pi) / (6) + 2 pi n#
However, the domain is given as #x in [0, 2 pi]#.
Therefore, the points of inflection occur at #x = (pi) / (6)#, #x = (5 pi) / (6)# and #x = (3 pi) / (2)#.
