If √3 cosθ+sinθ=√2,then general value of θ is
nπ+(−1)n π4−π3,n∈Z
Given equation:√3cosθ+sinθ=√2 ...(i)This is of form α cosθ+b sin θ=cWhere a=√3,b=1 and c=√2Let:a=r sin α and b=r cos α.Now,r=√a2+b2=√(√3)2+12=2And,tan α=ab⇒tan alpha=√31⇒tan α=tanπ3⇒α=π3Putting a=√3=r sin α and b=1=r cosα in equation (i),we get:r cosθ sinα+r sinθ cosα=√2⇒r sin(θ+α)=√2⇒2 sin(θ+α)=√2⇒sin(θ+π3)=1√2⇒sin(θ+π3)=cosπ4⇒θ+π3=nπ+(−1)n π4,n∈Z⇒θ=nπ+(−1)n π4−π3,n∈Z