Find general solution of sin^2 theta sec theta + root 3 tan theta =0 Share with your friends Share 2 Sanjay Kumar Manjhi answered this Dear Student,sin2θsecθ+3tanθ=0⇒sin2θcosθ+3tanθ=0⇒tanθsinθ+3tanθ=0⇒tanθ(sinθ+3)=0So,tanθ=0θ=nπ, where n∈Z .................(i)andsinθ+3=0⇒θ=sin-1(-3) General solution of above equation is;θ=2πn-sin-1(3), where n∈Z ................(ii)andθ=2πn+π+sin-1(3), where n∈Z .....................(iii)From equation (i), (ii) and (iii), we get;So, θ=nπ, 2πn-sin-1(3), 2πn+π+sin-1(3) where n∈ZRegards. 23 View Full Answer Sai Akash answered this no thought -16