Find general solution of sin^2 theta sec theta + root 3 tan theta =0

Dear Student,sin2θsecθ+3tanθ=0sin2θcosθ+3tanθ=0tanθsinθ+3tanθ=0tanθ(sinθ+3)=0So,tanθ=0θ=,  where nZ .................(i)andsinθ+3=0θ=sin-1(-3) General solution of above equation is;θ=2πn-sin-1(3), where nZ ................(ii)andθ=2πn+π+sin-1(3), where nZ .....................(iii)From equation (i), (ii) and (iii), we get;So, θ=, 2πn-sin-1(3), 2πn+π+sin-1(3)     where nZRegards.

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