The correct option is C nπ−π6≤k≤nπ+π6,n∈I
Given, sinx+cos(k+x)+cos(k−x)=2
This equation is of the form acosx+bsinx=c
Here a=2cosk,b=1 and c=2
Since for real solutions, |c|≤√a2+b2
∴|2|≤√1+4cos2k⇒2≤√1+4cos2k
⇒cos2k≥34⇒sin2k≤14
⇒sin2k=14≤0⇒[sink+12][sink−12]≤0
⇒−12≤sink≤12⇒nπ−π6≤k≤nπ+π6