1
$\begingroup$

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

Improve this question
$\endgroup$
1
$\begingroup$

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

Improve this answer
$\endgroup$
0
$\begingroup$

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

Improve this answer
$\endgroup$
  • $\begingroup$ i have just added. my silly mistake. thank you :D $\endgroup$ – sonarypt Dec 6 '16 at 14:29
0
$\begingroup$

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?

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.