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How you get X when: $\sin(x) + \tan(x) =0$ It's pretty basic but I just don't understand What I did until now is: $\tan(x) = -\sin(x) \Rightarrow \frac{\sin(x)}{\cos(x)} = -\sin(x)$

But after that I just don't know how to get to an end. Please help

Thanks

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Either you have $\sin x = 0$ or you can simplify the above expression to $$ - \cos x = 1 $$

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If $\sin x=0\Rightarrow x=2k\pi \ (k\in\mathbb{Z})$. Else, after eliminating $\sin x$ both side, it implies $\cos x=-1\rightarrow x=(2k+1)\pi \ (k\in\mathbb{Z})$

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  • $\begingroup$ i have just added. my silly mistake. thank you :D $\endgroup$ – sonarypt Dec 6 '16 at 14:29
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If $$\tan x + \sin x=0$$ then \begin{align} \frac{\sin x}{\cos x} + \sin x &= 0 \\ \iff \sin x + \sin x \cos x &=0 \\ \iff \sin x(1+\cos x) &=0 \end{align} Can you finish from here?

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