Question

solve each equation on the interval

$0\le \theta <2\pi$

$\tan \left(\theta \right)+\sqrt{3}=\frac{1}{\cos \left(\theta \right)}$

Short answer

$\theta =n\pi +{(-1)}^n\frac{\pi }{3}-\frac{\pi }{3}$

Step-by-step solution

Step 1.The information given istanθ+3=1cosθ

a trigonometric equation.

Step 2. Solving it we get,

$$\begin{gathered} \tan \left(\theta \right)+\sqrt{3}=\frac{1}{\cos \left(\theta \right)} \\ \frac{\sin \theta }{\cos \theta }+\sqrt{3}=\frac{1}{\cos \theta } \\ \frac{\sin \theta +\sqrt{3}\cos \theta }{\cos \theta }=\frac{1}{\cos \theta } \\ \sin \theta +\sqrt{3}\cos \theta =1 \end{gathered}$$

Step 3. Multiplying both the side by 12

$$\begin{gathered} \frac{1}{2}\sin \theta +\frac{\sqrt{3}}{2}\cos \theta =\frac{1}{2} \\ \cos \left(\frac{\pi }{3}\right)\sin \left(\theta \right)+\sin \left(\frac{\pi }{3}\right)\cos \left(\theta \right)=\sin \left(\frac{\pi }{3}\right) \\ \sin \left(\theta +\frac{\pi }{3}\right)=\sin \left(\frac{\pi }{3}\right) \end{gathered}$$

Step 4. The general solution we get is

$\theta =n\pi +{(-1)}^n\frac{\pi }{3}-\frac{\pi }{3}$