Answer to Question #155165 in Trigonometry for Daniel Romanczuk

Question

Solve for θ to the nearest hundredth, where 0 ≤ x ≤ 2π. a) 3 tan2 θ – 2 tan θ = 1

Answer 1

Solution

• First write the $\tan2\theta$ in terms of $\tan\theta$ to get the equation in like terms.
• Then,

$\qquad\qquad \begin{aligned} 3\bigg(\frac{2\tan\theta}{1-\tan^2\theta}\bigg)-2\tan\theta&=1\\ \end{aligned}$

• For easy recognition, substitute $\tan\theta=x$ & simplify the equation to find the roots of it.
• Then,

$\qquad\qquad \begin{aligned} \frac{6x}{1-x^2}-2x&=0\\ \frac{2x^3+x^2+4x-1}{1-x^2}&=0 \end{aligned}$

• Now $x \neq \pm1$ for the equation to exist.
• Then its about finding the roots of the numerator equation.
• They are

$x=\begin{cases} 0.2306\\ -0.365+1.427i\\ -0.365-1.426i \end{cases}$

• All those values are acceptable only for an algebraic variable like x but only 0.2306 is valid for $\tan \theta$ as $-\infty<\tan\theta<+\infty$
• Therefore,

$\qquad\qquad \begin{aligned} x&=\tan\theta=0.2306\\ \theta&= 0.2266 rad \end{aligned}$