Question

find all solutions to the following equation.
8 sin θ + 1 = 6 sin θ
select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. θ = (simplify your answer. type your answer(s) as an expression, using n as the variable, in the form a + bn where
0 ≤ a < 2π. type any angle measures in radians, using π as needed. use integers or fractions for any numbers
in the expression. use a comma to separate answers as needed.)
b. there is no solution.

Explanation:

Step1: Simplify the equation

Given \(8\sin\theta + 1=6\sin\theta\).
Subtract \(6\sin\theta\) from both sides:
\(8\sin\theta-6\sin\theta + 1 = 0\).
\(2\sin\theta+1 = 0\).
Then \(2\sin\theta=-1\), so \(\sin\theta=-\frac{1}{2}\).

Step2: Find the general solution of \(\sin\theta =-\frac{1}{2}\)

We know that \(\sin\theta =-\frac{1}{2}\) has solutions \(\theta=\frac{7\pi}{6}+2n\pi\) and \(\theta=\frac{11\pi}{6}+2n\pi\), where \(n\in\mathbb{Z}\) (because \(\sin\theta\) has a period of \(2\pi\) and \(\sin\frac{7\pi}{6}=-\frac{1}{2}\), \(\sin\frac{11\pi}{6}=-\frac{1}{2}\)).
When \(n = 0\), for \(0\leq a<2\pi\), the solutions are \(\theta=\frac{7\pi}{6},\frac{11\pi}{6}\).

Answer:

A. \(\theta=\frac{7\pi}{6}+2n\pi,\frac{11\pi}{6}+2n\pi\) (where \(n\in\mathbb{Z}\))