Question

3 tan^3θ = tan θ Solve for theta

Answer 1

start with
\[3\tan^3(\theta)-\tan(\theta)=0\]then
\[\tan(\theta)(3\tan^2(\theta)-1)=0\] so
\[\tan(\theta)=0\] or
\[3\tan^2(\theta)-1=0\implies \tan(\theta)=\pm\frac{1}{\sqrt{3}}\]\]

Answer 2

to solve for \(\theta\) if \(\tan(\theta)=0\) then \(\theta =0,\pi, 2\pi, ...\)

Answer 3

\[\tan(\theta)=\frac{1}{\sqrt{3}}\implies \theta =\frac{\pi}{6}\] and since tangent is periodic with period \(\pi\) you can add any multiple of \(\pi\)

Answer 4

your last job is to solve
\[\tan(\theta)=-\frac{1}{\sqrt{3}}\] and since tangent is odd that will give \(-\frac{\pi}{6}\) etc