The correct option is D x=(2n+1)π2,n∈Z
Since, |a+b|=|a|+|b|⇒ab≥0
So, |sinx+cosx|=|sinx|+|cosx|
⇒sinxcosx≥0
Possible cases are :
(i) sinx and cosx are of same sign.
i.e., x∈I quadrant or x∈III quadrant.
(ii) sinx=0⇒x=nπ,n∈Z
(iii) cosx=0⇒x=(2n+1)π2,n∈Z