If  16sinθ,cosθ and  tanθ  are in GP, and range of θ is 0,nπ and  has 5 values, then n =

Since, 16sinθ,cosθ,tanθ are in G.P.∴cos2θ=16sinθ.tanθcos2θ =16sin2θcosθ  put sin2θ =1-cos2θ⇒6cos3θ+cos2θ−1=0Note that  cosθ=12 satisfies the equation (by trial)∴(2cosθ−1)(3cos2θ+2cosθ+1)=0⇒cosθ=12           (other cosθ values of  are imaginary)⇒cosθ=cosπ3⇒θ=2nπ±π3,n∈Z⇒for n=0, θ=±π3     ∴θ=π3,2π-π3,2π+π3,4π-π3 and 4π+π3are the 5 values taken by θ in 0,5π ⇒n=5   for  n=1, θ=2π±π3     for   n=2, θ=4π±π3