Trigonometric Equations, Trouble solving. $\displaystyle \sin x + \sqrt3\cos x =0$
$0< x <360^\circ$
Hey, I haven't done this question for a while and I forgot how to solve it. I have made attempts to solve it, such as splitting the sinx and cosx, but I still cant figure it out. Could some one please hint me or walk me through the steps?
Thank you.
 A: Since
$$\sin x+\sqrt 3\cos x=2(\sin x\times(1/2)+\cos x\times (\sqrt 3/2))=2\sin (x+60^\circ),$$
you'll have$$\sin x+\sqrt 3\cos x=0\iff \sin(x+60^\circ)=0.$$
Since
$$0\lt x\lt 360^\circ\iff 60^\circ \lt x+60^\circ\lt 420^\circ,$$
you'll have$$x+60^\circ=180^\circ, 360^\circ.$$
Hence, 
$$x=120^\circ, 300^\circ.$$
A: $$\sin x+\sqrt3\cos x=0\iff \sin x=-\sqrt3\cos x$$
$$\iff \tan x=-\sqrt3=-\tan60^\circ=\tan(-60^\circ)$$ as $\tan(-y)=-\tan y$
$$x=n180^\circ+(-60^\circ)$$ where $n$ is any integer
A: Let $x$ denote a fixed but arbitrary element of $(0,2\pi)$.
Now the following are equivalent.


*

*$\sin x + \sqrt{3} \cos x = 0$

*$\sin x = -\sqrt{3} \cos x$

*Either $\tan x = -\sqrt{3},$ or both $\cos x = 0$ and $\sin x = 0$


You can probably take it from here; please comment if you need a bit more help.
A: $\sin x + \sqrt{3}\cos x =0$ can be written as $(e^{ix}-e^{-ix})/2i+(e^{ix}+e^{-ix})\sqrt{3}/2=0$. This simplifies to $e^{2ix}=(\frac12-i\frac{\sqrt3}2)^2$. This is the same as $e^{2ix}=(e^{-\pi i/3})^2$ or $e^{2ix}=e^{-2\pi i/3}$, so $2x=-2\pi/3 + 2\pi k$, i.e., $x=-\pi/3+\pi k$ for integral $k$.
