<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] 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 $sin\theta$.
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