\[ \text { If } 8 \sin x=4+\cos x \text {, the values of } \sin x \text { are: } \]

To find the values of \(\sin x\) given the equation \(8\sin x = 4 + \cos x\), we can use trigonometric identities to rewrite the equation in terms of a single trigonometric function.

The given equation is:
\[8\sin x = 4 + \cos x\]

We know the Pythagorean identity: \(\sin^2 x + \cos^2 x = 1\). Our goal is to express \(\cos x\) in terms of \(\sin x\) (or vice versa) to solve for \(\sin x\). Since the equation involves both \(\sin x\) and \(\cos x\), and we’re looking to find \(\sin x\), let’s isolate \(\cos x\) and then use the Pythagorean identity.

Rearrange the given equation to isolate \(\cos x\):
\[\cos x = 8\sin x – 4\]

Using the Pythagorean identity \(\cos^2 x = 1 – \sin^2 x\), we substitute for \(\cos x\) in the equation:
\[1 – \sin^2 x = (8\sin x – 4)^2\]

Expanding the right side and moving all terms to one side gives us a quadratic equation in \(\sin x\):
\[1 – \sin^2 x = 64\sin^2 x – 64\sin x + 16\]

Combine like terms:
\[65\sin^2 x – 64\sin x + 15 = 0\]

This is a quadratic equation in \(\sin x\). To solve for \(\sin x\), we use the quadratic formula where \(a = 65\), \(b = -64\), and \(c = 15\):
\[\sin x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\]
\[= \frac{64 \pm \sqrt{(-64)^2 – 4 \cdot 65 \cdot 15}}{2 \cdot 65}\]
\[= \frac{64 \pm \sqrt{4096 – 3900}}{130}\]
\[= \frac{64 \pm \sqrt{196}}{130}\]
\[= \frac{64 \pm 14}{130}\]

So, we have two possible solutions for \(\sin x\):
1. \(\sin x = \frac{64 + 14}{130} = \frac{78}{130} = \frac{39}{65} = \frac{3}{5}\)
2. \(\sin x = \frac{64 – 14}{130} = \frac{50}{130} = \frac{25}{65} = \frac{5}{13}\)

Therefore, the values of \(\sin x\) that satisfy the given equation are \(\frac{3}{5}\) and \(\frac{5}{13}\).