Question

solve each equation on the interval $0\le \theta <2\pi$

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

Short answer

$\theta =(n\pi +\frac{\mathrm{\pi }}{3})+{(-1)}^{\left(n\right)}\frac{\mathrm{\pi }}{6}$

Step-by-step solution

Step 1.Given information is

a trignometric equation$\sin \theta -\sqrt{3}\cos \theta =1$

Step 2.multiplying both the side by 12, we get12sinθ-32cosθ=12cosπ3sinθ-sinπ3cosθ=sinπ6sinθ-π3=sinπ6

we know that

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

Step 3.the general solution we get is

$\theta =n\pi +{(-1)}^n\frac{\pi }{6}+\frac{\pi }{3}$