7tanθ + tanθ - 2 $\sqrt{\ 3}$ = 0
You asked:
Solve the equation \(7 \cdot \tan\left( θ \right) + \tan\left( θ \right) - 2 \cdot \sqrt{3} = 0\) for the variable \(θ\).
MathBot Answer:
The principal value in \(- \frac{\pi}{2} < θ < \frac{\pi}{2}\) is \(θ \approx 0.40863786\)
The real solutions are: \[\begin{aligned}θ &= \left\{\frac{\arctan{\left(\frac{8 \sqrt{3}}{13} \right)}}{2} + 2 \pi n\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 0.40863786\; \middle|\; n \in \mathbb{Z}\right\}\\θ &= \left\{\pi + \frac{\arctan{\left(\frac{8 \sqrt{3}}{13} \right)}}{2} + 2 \pi n\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 3.5502305\; \middle|\; n \in \mathbb{Z}\right\}\end{aligned}\]