Let the given by #y=sin x+cos^2 x#
Determine the first derivative #dy/dx# then equate to zero, that is #dy/dx=0#
Let us begin
from the given
#y=sin x+cos^2 x=sin x+ (cos x)^2#
#d/dx(y)=d/dx(sin x) + d/dx(cos x)^2#
#dy/dx=cos x * dx/dx+ 2*(cos x)^((2-1))*d/dx(cos x)#
#dy/dx=cos x * 1 +2*(cos x)^1*(-sin x)*dx/dx#
#dy/dx=cos x-2*sin x*cos x*1#
#dy/dx=cos x-2*sin x*cos x#
Equate #dy/dx=0#
#cos x-2*sin x*cos x=0#
solve by factoring
#cos x(1-2 sin x)=0#
Equate each factor to zero
#cos x=0" " "# the first factor
#arccos(cos x)=arccos 0#
#x=pi/2#
find #y#, using the original equation
#y=sin x+cos^2 x#
#y=sin (pi/2)+cos^2 (pi/2)#
#y=1+(0)^2#
#y=1#
solution #(pi/2, 1)=(1.57, 1)" "#the Minimum Point
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#1-2 sin x=0" " " "# the second factor
#2*sin x=1#
#sin x=1/2#
#arcsin(sin x)=arcsin (1/2)#
#x=pi/6# also #x=(5pi)/6#
find #y#, using #x=pi/6# in the original equation
#y=sin x+cos^2 x#
#y=sin (pi/6)+cos^2 (pi/6)#
#y=1/2+(sqrt3/2)^2#
#y=1/2+3/4#
#y=5/4#
solution #(pi/6, 5/4)=(0.523599, 1.25)" " "#the Maximum Point
the other Maximum Point is at #((5pi)/6, 5/4)=(2.61799, 1.25)#
because #sin (pi/6)=sin ((5pi)/6)#. That is why there are two maximum points.
Kindly see the graph and locate the critical points
graph{y=sin x+(cos x)^2 [-1, 5, -1, 1.5]}
God bless....I hope the explanation is useful.
