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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 !

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    $\begingroup$ If $u=\cos\theta$, you got the equation $65u^2 - 8u - 48 = 0$. You can solve it, get $\cos\theta$, and then find $\sin\theta$. $\endgroup$ – Guy Jul 9 '16 at 7:29
  • $\begingroup$ Ain't there a shorter way @Guy $\endgroup$ – Amritanshu Jul 9 '16 at 7:31
  • $\begingroup$ @Amritanshu: You have a sign mistake at the very last step. $\endgroup$ – mathlove Jul 9 '16 at 7:32
  • $\begingroup$ Corrected it @mathlove $\endgroup$ – Amritanshu Jul 9 '16 at 7:33
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    $\begingroup$ You can solve $a\sin x+b\cos x=c$ in general. math.stackexchange.com/questions/213545/… $\endgroup$ – user236182 Jul 9 '16 at 7:41
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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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$$8\sin(\theta)- \cos(\theta) =4$$ 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 $$\sin(\theta)\cos(\alpha)-\cos(\theta)\sin(\alpha) = \frac{4}{\sqrt{65}}$$ $$\sin(\theta -\alpha) = \frac{4}{\sqrt{65}}$$ $$\theta = \sin^{-1}(\frac{4}{\sqrt{65}}) + \alpha $$ $$= \sin^{-1}(\frac{4}{\sqrt{65}}) + \sin^{-1}(\frac{1}{\sqrt{65}})$$

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$$8\sin\theta=4+\cos\theta$$ $$8\sin\theta-\cos\theta=4$$ $$\sqrt{65}\sin\alpha\sin\theta-\sqrt{65}\cos\alpha\cos\theta=4$$ where $\alpha=\tan^{-1}8$ $$-\sqrt{65}\cos(\theta+\alpha)=4$$ Can you continue from here?

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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 $\cos(\frac{\theta}{2}) = 0 \Leftrightarrow \theta = \pi [2\pi] \Rightarrow \sin(\theta) = 0$

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  • $\begingroup$ You first have to prove that $\tan\frac{\theta}{2}$ exists, i.e. that $\cos \frac{\theta}{2}\neq 0$, i.e. that $\frac{\theta}{2}\neq (2k+1)\frac{\pi}{2}$, $k\in\mathbb Z$, which is simple to prove. $\endgroup$ – user236182 Jul 9 '16 at 7:54
  • $\begingroup$ It should be $\Leftrightarrow \theta=\pi(2k+1)$, $k\in\mathbb Z$. $\endgroup$ – user236182 Jul 9 '16 at 7:57
  • $\begingroup$ @user236182: $\theta = \pi [2\pi] \Leftrightarrow \theta = \pi + 2 k \pi, k \in \mathbb{Z} \Leftrightarrow \theta = \pi(2k+1), k \in \mathbb{Z}$ $\endgroup$ – Faibbus Jul 9 '16 at 8:39
  • $\begingroup$ So you use $[2\pi]$ to denote $\bmod 2\pi$. I haven't seen this notation before. I thought you meant $\pi[2\pi]=\pi(2\pi)=\pi \cdot 2\pi$. $\endgroup$ – user236182 Jul 9 '16 at 8:42

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