1 - First we have to find the second derivative of our function.
2 - Second, we equate that derivative#((d^2y)/(dx^2))# to zero
#y=sinx +cosx#
#=>(dy)/(dx)=cosx-sinx#
#=>(d^2y)/(dx^2)=-sinx-cosx#
Next, #-sinx-cosx=0#
#=>sinx+cosx=0#
Now, we shall express that in the form #Rcos(x+lamda)#
Where #lambda# is just an acute angle and #R# is a positive integer to be determined. Like this
#sinx+cosx=Rcos(x+lambda)#
#=>sinx+cosx=Rcosxcoslamda - sinxsinlamda#
By equating the coefficients of #sinx# and #cosx# on either side of the equation,
#=>Rcoslamda=1#
and #Rsinlambda=-1#
#(Rsinlambda)/(Rcoslambda)=(-1)/1=>tanlambda=-1=>lambda=tan^-1(-1)=-pi/4#
And #(Rcoslambda)^2+(Rsinlambda)^2=(1)^2+(-1)^2#
#=>R^2(cos^2x+sin^2x)=2#
But we know the identity , #cos^2x+sin^2=1#
Hence, #R^2(1)=2=>R=sqrt(2)#
In a nut shell, #(d^2y)/(dx^2)=-sinx-cosx=sqrt(2)cos(x-pi/4)=0#
#=>sqrt(2)cos(x-pi/4)=0#
#=>cos(x-pi/4)=0=cos(pi/2)#
So the general solution of #x# is : #x-pi/4=+-pi/2+2kpi# , #kinZZ#
#=>x=pi/4+-pi/2+2kpi#
So the points of inflexion will be any point that has coordinates :
#(pi/4+-pi/2+2kpi , sqrt(2)cos(pi/4+-pi/2-pi/4))#
We have two cases to deall with,
Case 1
#(pi/4+pi/2+2kpi , sqrt(2)cos(pi/4+pi/2-pi/4))#
#=>((3pi)/4+2kpi , sqrt(2)cos(pi/2))#
#=>((3pi)/4+2kpi , 0)#
Case 2
#(pi/4-pi/2+2kpi , sqrt(2)cos(pi/4-pi/2-pi/4))#
#=>(-pi/2+2kpi , sqrt(2)cos(-pi/2))#
#=>((-pi/2+2kpi , 0))#
