Solution
First write the tan 2 θ \tan2\theta tan 2 θ in terms of tan θ \tan\theta tan θ to get the equation in like terms. Then, 3 ( 2 tan θ 1 − tan 2 θ ) − 2 tan θ = 1 \qquad\qquad
\begin{aligned}
3\bigg(\frac{2\tan\theta}{1-\tan^2\theta}\bigg)-2\tan\theta&=1\\
\end{aligned} 3 ( 1 − tan 2 θ 2 tan θ ) − 2 tan θ = 1
For easy recognition, substitute tan θ = x \tan\theta=x tan θ = x & simplify the equation to find the roots of it. Then, 6 x 1 − x 2 − 2 x = 0 2 x 3 + x 2 + 4 x − 1 1 − x 2 = 0 \qquad\qquad
\begin{aligned}
\frac{6x}{1-x^2}-2x&=0\\
\frac{2x^3+x^2+4x-1}{1-x^2}&=0
\end{aligned} 1 − x 2 6 x − 2 x 1 − x 2 2 x 3 + x 2 + 4 x − 1 = 0 = 0
Now x ≠ ± 1 x \neq \pm1 x = ± 1 for the equation to exist. Then its about finding the roots of the numerator equation. They are x = { 0.2306 − 0.365 + 1.427 i − 0.365 − 1.426 i x=\begin{cases}
0.2306\\
-0.365+1.427i\\
-0.365-1.426i
\end{cases} x = ⎩ ⎨ ⎧ 0.2306 − 0.365 + 1.427 i − 0.365 − 1.426 i
All those values are acceptable only for an algebraic variable like x but only 0.2306 is valid for tan θ \tan \theta tan θ as − ∞ < tan θ < + ∞ -\infty<\tan\theta<+\infty − ∞ < tan θ < + ∞ Therefore, x = tan θ = 0.2306 θ = 0.2266 r a d \qquad\qquad
\begin{aligned}
x&=\tan\theta=0.2306\\
\theta&= 0.2266 rad
\end{aligned} x θ = tan θ = 0.2306 = 0.2266 r a d
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