Find the critical points and the local extreme values. f(x)=(sinx)^2-[sqrt(3)*sinx] 0

To determine the local extreme values of the function,
we'll have to do the first derivative test.

For this
reason, we'll determine the expression of the first
derivative:

f'(x) =
{(sinx)^2-[sqrt(3)*sinx]}'

f'(x) = 2sin x*cos x - sqrt3*cos
x

Now, to calculate the local extreme of the function,
we'll determine the roots of the first derivative:

 f'(x) =
0

2sin x*cos x - sqrt3*cos x =
0

We'll factorize by cos
x:

cos x(2sin x - sqrt 3) =
0

We'll put each factor as
zero:

cos x = 0

This in an
elementary equation:

x = arccos
0

x =
pi/2

2sin x - sqrt 3 =
0

2sin x = sqrt3

sin x = sqrt
3/2

x = arcsin (sqrt
3/2)

x =
pi/3

x = pi -
pi/3

x =
2pi/3

Critical values of x:
{pi/3 , pi/2 , 2pi/3}.

The
local extremes of the function are the points whose x coordinate has the values:{pi/3 ,
pi/2 , 2pi/3}.

Now, we'll
determine the local extremes. For this reason, we'll substitute x by the critical
values:

f(pi/3) = (sin pi/3)^2 - sqrt3*sin
pi/3

f(pi/3) = 3/4 -
3/2

f(pi/3)
=-3/4

f(pi/2)
=  (sin pi/2)^2 - sqrt3*sin
pi/2

f(pi/2)
= 1 - sqrt
3

f(2pi/3)
= (sin 2pi/3)^2 - sqrt3*sin
2pi/3

f(2pi/3)
= f(pi/3) =
-3/4