Question

In Problems 11–34, solve each equation on the interval $0\le \theta \le 2\pi$

$\tan \frac{\theta }{2}=\sqrt{3}$

Short answer

The solution set is$\left\{\frac{2\mathrm{\pi }}{3}\right\}$

Step-by-step solution

Step 1. Given Information 

In the given problem we have to solve each equation on the interval$0\le \theta \le 2\pi$.

$\tan \frac{\theta }{2}=\sqrt{3}$

Step 2. In the interval [0,2π), the tangent function equals 3 at π3.

So, we know that $\frac{\theta }{2}$must equal $\frac{\mathrm{\pi }}{3}$.

To find these solutions, write the general formula that gives all the solutions.

$\frac{\theta }{2}=\frac{\mathrm{\pi }}{3}+\mathrm{\pi n}$

Multiply with 2 on both side

$$\begin{gathered} \frac{\theta }{2}\times 2=2\times \left(\frac{\mathrm{\pi }}{3}+\mathrm{\pi n}\right) \\ \theta =\frac{2\mathrm{\pi }}{3}+2\mathrm{\pi n} \end{gathered}$$

Step 3. The general formula is θ=2π3+2πn.

So the value of given function in interval $[0,2\mathrm{\pi })$is

$\theta =\frac{2\mathrm{\pi }}{3}+2\mathrm{\pi n}$

role="math" localid="1646586437011" $$\begin{gathered} \theta =\frac{2\mathrm{\pi }}{3}+2\pi \times 0 \\ \theta =\frac{2\mathrm{\pi }}{3} \end{gathered}$$

So the solution set is$\left\{\frac{2\mathrm{\pi }}{3}\right\}$.