Answer:
[tex]\large\boxed{x=\dfrac{\pi}{6}+2k\pi\ \vee\ x=\dfrac{5\pi}{6}+2k\pi,\ k\in\mathbb{Z}}[/tex]
Step-by-step explanation:
[tex]\sin x-(3\sin x-1)=0\\\\\sin x-3\sin x+1=0\qquad\text{subtract 1 from both sides}\\\\-2\sin x=-1\qquad\text{divide both sides by 2}\\\\\sin x=\dfrac{1}{2}\Rightarrow x=\dfrac{\pi}{6}+2k\pi\ \vee\ x=\dfrac{5\pi}{6}+2k\pi,\ k\in\mathbb{Z}[/tex]
[tex]\text{Equation:}\\\\\sin x=a\\\\\text{has solutions}\\\\x=\theta+2k\pi\ \vee\ x=(\pi-\theta)+2k\pi\\\\\text{Why}\ 2k\pi?\\\text{Because the sine function has a period of}\ 2\pi.[/tex]
[tex]\text{look at the table}\\\\\sin x=\dfrac{1}{2}\to x=\dfrac{\pi}{6}\ \vee\ x=\pi-\dfrac{\pi}{6}=\dfrac{5\pi}{6}[/tex]
[tex]\text{Other solution:}\\\\\sin x=\dfrac{1}{2}\Rightarrow x=\sin^{-1}\dfrac{1}{2}\\\\x=\dfrac{\pi}{6}+2k\pi\ \vee\ x=\dfrac{5\pi}{6}+2k\pi[/tex]
