Question

Please help! Solve the equation on the interval [0,2pi)

Answer 1

Answer:

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

Step-by-step explanation:

$\begin{aligned}3 \sin x & = \sin x + 1\\\implies 3 \sin x - \sin x & = 1\\2 \sin x & = 1\\\sin x & = \dfrac{1}{2}\\\implies x & = \sin^{-1}\left(\dfrac{1}{2}\right)\\x & = \dfrac{\pi}{6} \pm 2 \pi n, \dfrac{5 \pi}{6}\pm 2 \pi n\end{aligned}$

Therefore, the value of x in the interval [0, 2π) is:

$x = \dfrac{\pi}{6}, \dfrac{\boxed{5} \pi}{\boxed{6}}$

Answer 2

Answer:

$\frac{5pi}{6}$

Step-by-step explanation: