- #36

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I get $$ \tan \theta (\sin^2 \theta ) = \dfrac{q^2}{8 \pi \epsilon_0 mgl^2} $$which is the same as $$\frac{ \tan^3 \theta}{1+ \tan^2 \theta} = \dfrac{q^2}{8 \pi \epsilon_0 mgl^2} $$Answer provided that I just found at the back of the book $$ \tan^3 \theta ( 1+ \tan^2 \theta ) = \dfrac{q^2}{8 \pi \epsilon_0 mgl^2} $$