Question

Find the general solution
tan^2theta=square root 3 tan theta

Answer 1

\[\tan ^{2}\theta=\sqrt{3}\tan \theta\]

Answer 2

Can you repeat how you did the last problem? didn't you substitute for that one?

Answer 3

You have:
\[\tan^2 \psi-\sqrt3 \tan \psi=0; \lambda = \tan \psi \implies \lambda^2-\sqrt3 \lambda=0 \implies \lambda= \pm \sqrt[4]{3}\]\[\implies \tan \psi = \pm \sqrt[4]{3} \implies \psi = \arctan(\pm \sqrt[4]{3})\]
Which you can plug into wolfram to get ~53 degrees