The number of values of θ in  [0,3π] satisfying sin3θ+cos3θ+3sinθcosθ-1=0 is

We have  sin3θ+cos3θ−1=−3sinθcosθ⇒sin3θ+cos3θ+(-1)3=3(sinθ)(cosθ)(−1)which is in the form       a3+b3+c3=3abc⇒∴sinθ=cosθ=−1  which is impossibleor   sinθ+cosθ+(−1)=0⇒sinθ+cosθ=1⇒12sinθ+12cosθ=12⇒cosθ−π4=cosπ4⇒θ−π4=2nπ±π4⇒θ=2nπ+π2     or     θ=  2nπ⇒ for n=0,θ=π2,0      and    for   n=1,    θ=5π2,2π