solve the equation on the interval $0,2\\pi)$. \n\n$3\\sin x = \\sin x + 1$\n\n$x = \\frac{\\pi}{?},\\frac{\\…

Explanation:

Step1: Simplify the equation

Subtract $\sin x$ from both sides: $$3\sin x-\sin x=\sin x + 1-\sin x$$ $$2\sin x=1$$

Step2: Solve for $\sin x$

Divide both sides by 2: $$\sin x=\frac{1}{2}$$

Step3: Find $x$ in the interval $[0,2\pi)$

We know that $\sin x=\frac{1}{2}$ when $x = \frac{\pi}{6}$ (since $\sin\frac{\pi}{6}=\frac{1}{2}$) and $x=\pi-\frac{\pi}{6}=\frac{5\pi}{6}$ (using the identity $\sin(\pi - \alpha)=\sin\alpha$) in the interval $[0,2\pi)$.

Answer:

$x=\frac{\pi}{6},\frac{5\pi}{6}$