How do you find the solution to $3 {tan}^{2} θ = 1$ $3tan^2theta=1$ if $0 \leq θ < 360$ $0<=theta<360$ ?

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Answer 1 · 2018-06-13T21:01:50

$θ = 30, 210, 150, 330$ $\theta=30,210,150,330$

${tan}^{2} θ = \frac{1}{3}$ $tan^{2}\theta = \frac{1}{3}$

Therefore:

$tan θ = \pm \frac{1}{\sqrt{3}}$ $tan\theta=+-\frac{1}{\sqrt{3}}$

Solutions cna be found such that:

$θ = {tan}^{- 1} (\frac{1}{\sqrt{3}}) = 30, 210$ $\theta=tan^{-1}(\frac{1}{\sqrt{3}})=30, 210$ Using the 180 cycle in a tan graph.

doing the same for the negation version gives us:
$\theta=tan^{-1}(-\frac{1}{\sqrt{3}})=cancel(-30),150,330$

therefore all the solutions are:
$\theta=30,210,150,330$

How do you find the solution to 3tan^2theta=13tan2θ=13tan^2theta=1 if 0<=theta<3600≤θ<3600<=theta<360?