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 !
 A: $$8\sin(\theta)- \cos(\theta) =4$$
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 
$$\sin(\theta)\cos(\alpha)-\cos(\theta)\sin(\alpha) = \frac{4}{\sqrt{65}}$$
$$\sin(\theta -\alpha) = \frac{4}{\sqrt{65}}$$
$$\theta = \sin^{-1}(\frac{4}{\sqrt{65}}) + \alpha $$
$$= \sin^{-1}(\frac{4}{\sqrt{65}}) + \sin^{-1}(\frac{1}{\sqrt{65}})$$
A: Replacing $cos\theta=\pm\sqrt{1-sin^2\theta}$ in your equation, $8\sin\theta = 4 + \cos\theta$, and considering $u=sin\theta$, we have
$8u=4\pm\sqrt{1-u^2}$ which leads to $64u^2 +16 - 64u=1-u^2\to 65u^2-64u+15=0$.
Now, you only need to solve this equation to find $u$ that is $sin\theta$.
A: $$8\sin\theta=4+\cos\theta$$
$$8\sin\theta-\cos\theta=4$$
$$\sqrt{65}\sin\alpha\sin\theta-\sqrt{65}\cos\alpha\cos\theta=4$$
where $\alpha=\tan^{-1}8$
$$-\sqrt{65}\cos(\theta+\alpha)=4$$
Can you continue from here?
A: Or you can use the half angle formulas:
$\cos(\theta) = \frac{1 - \tan^2(\frac{\theta}{2})}{1 + \tan^2(\frac{\theta}{2})} $
$\sin(\theta) = \frac{2 \tan(\frac{\theta}{2})}{1 + \tan^2(\frac{\theta}{2})}$
So as to get:
$ 8 \frac{2 \tan(\frac{\theta}{2})}{1 + \tan^2(\frac{\theta}{2})} = 4 + \frac{1 - \tan^2(\frac{\theta}{2})}{1 + \tan^2(\frac{\theta}{2})} $
$ 3 \tan^2(\frac{\theta}{2}) - 16 \tan(\frac{\theta}{2}) +5 = 0$
Which you can solve for $\tan(\frac{\theta}{2})$ that you can then use to compute $\sin(\theta)$
Edit: I forgot to mention that you can do this only because $\cos(\frac{\theta}{2}) \neq 0$, because $\cos(\frac{\theta}{2}) = 0 \Leftrightarrow \theta = \pi [2\pi] \Rightarrow \sin(\theta) = 0$
