Find $\sin \theta $ in the equation $8\sin\theta = 4 + \cos\theta$ Find $\sin\theta$ in the following trigonometric equation
$8\sin\theta = 4 + \cos\theta$
My try ->
$8\sin\theta = 4 + \cos\theta$
[Squaring Both the Sides]
=> $64\sin^{2}\theta = 16 + 8\cos\theta + \cos^{2}\theta$
=> $64\sin^{2}\theta - \cos^{2}\theta= 16 + 8\cos\theta $
[Adding  on both the sides]
=> $64\sin^{2}\theta + 64\cos^{2}\theta= 16 + 8\cos\theta + 65\cos^{2}\theta$
=> $64 = 16 + 8\cos\theta + 65\cos^{2}\theta$
=> $48 = 8\cos\theta + 65\cos^{2}\theta$
=> $48 = \cos\theta(65\cos\theta + 8)$
I can't figure out what to do next !

$$8\sin(\theta)- \cos(\theta) =4$$
Dividing by $\sqrt{1^2 +8^2}$ 
$$\frac{8\sin(\theta)}{\sqrt{65}}- \frac{\cos(\theta)}{\sqrt{65}} =\frac{4}{\sqrt{65}}$$
Let $\sin(\alpha) = \frac{1}{\sqrt{65}}$, we $\cos(\alpha) = \frac{8}{\sqrt{65}}$.
Thus 
$$\sin(\theta)\cos(\alpha)-\cos(\theta)\sin(\alpha) = \frac{4}{\sqrt{65}}$$
$$\sin(\theta -\alpha) = \frac{4}{\sqrt{65}}$$
$$\theta = \sin^{-1}(\frac{4}{\sqrt{65}}) + \alpha $$
$$= \sin^{-1}(\frac{4}{\sqrt{65}}) + \sin^{-1}(\frac{1}{\sqrt{65}})$$
