168k views
3 votes
Please help! Solve the equation on the interval [0,2pi)

Please help! Solve the equation on the interval [0,2pi)-example-1
User Blekione
by
4.7k points

2 Answers

6 votes

Answer:


(5pi)/(6)

Explanation:

User Casey Plummer
by
4.5k points
12 votes

Answer:


x = (\pi)/(6), (5 \pi)/(6)

Explanation:


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

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


x = (\pi)/(6), \frac{\boxed{5} \pi}{\boxed{6}}

User Fpointbin
by
4.5k points