Question

solve each equation on the interval

$\sqrt{3}\sin \theta +\cos \theta =1$

Short answer

$\theta =n\pi +{(-1)}^n\frac{\pi }{6}-\frac{\pi }{6}$

Step-by-step solution

Step 1.Given information is

a trignometirc equation$\sqrt{3}\sin \theta +\cos \theta =1$

Step 2. Multiplying both the side by 12

$$\begin{gathered} \frac{\sqrt{3}}{2}\sin \theta +\frac{1}{2}\cos \theta =\frac{1}{2} \\ \cos \left(\frac{\pi }{6}\right)\sin \theta +\sin \left(\frac{\pi }{6}\right)\cos \theta =\sin \left(\frac{\pi }{6}\right) \\ \sin \left(\theta +\frac{\pi }{6}\right)=\sin \left(\frac{\pi }{6}\right) \end{gathered}$$

Step 3.The general solution we get is

$\theta =n\pi +{(-1)}^n\frac{\pi }{6}-\frac{\pi }{6}$