find all solutions to the following equation.\n8 sin θ + 1 = 6 sin θ\nselect the correct choice below and…

find all solutions to the following equation.\n8 sin θ + 1 = 6 sin θ\nselect the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. θ = (simplify your answer. type your answer(s) as an expression, using n as the variable, in the form a + bn where\n0 ≤ a < 2π. type any angle measures in radians, using π as needed. use integers or fractions for any numbers\nin the expression. use a comma to separate answers as needed.)\nb. there is no solution.

find all solutions to the following equation.\n8 sin θ + 1 = 6 sin θ\nselect the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. θ = (simplify your answer. type your answer(s) as an expression, using n as the variable, in the form a + bn where\n0 ≤ a < 2π. type any angle measures in radians, using π as needed. use integers or fractions for any numbers\nin the expression. use a comma to separate answers as needed.)\nb. there is no solution.

Answer

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}))