<s>[INST] 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 !
 [/INST] 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 $\cos(\frac{\theta}{2}) = 0 \Leftrightarrow \theta = \pi [2\pi] \Rightarrow \sin(\theta) = 0$
</s>