Trigonometric Equations, Trouble solving.

Question

$\displaystyle \sin x + \sqrt3\cos x =0$

$0< x <360^\circ$

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.

score 3 · Answer 1 · answered Jan 20 '14 at 05:24

3

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, $$x=120^\circ, 300^\circ.$$

answered Jan 20 '14 at 05:24

mathlove

139,939

Clubbing $A\sin x+B\cos x=C$ is worth only if $C\ne0,$ right? – lab bhattacharjee Jan 20 '14 at 05:26
Well, I think this way can be used in the case $C=0.$ – mathlove Jan 20 '14 at 05:34
luckily here $\arccos \frac A{\sqrt{A^2+B^2}}$ is simple, else we had to deal with a little ugly trig inverse functions – lab bhattacharjee Jan 20 '14 at 05:36
Yeah, I agree with the point. – mathlove Jan 20 '14 at 05:38

score 2 · Answer 2 · answered Jan 20 '14 at 05:18

2

$$\sin x+\sqrt3\cos x=0\iff \sin x=-\sqrt3\cos x$$

$$\iff \tan x=-\sqrt3=-\tan60^\circ=\tan(-60^\circ)$$ as $\tan(-y)=-\tan y$

$$x=n180^\circ+(-60^\circ)$$ where $n$ is any integer

answered Jan 20 '14 at 05:18

lab bhattacharjee

274,582

Haven't you omitted the case $\cos (x) = 0$? – goblin GONE Jan 20 '14 at 05:20
@user18921, find my comment in your answer – lab bhattacharjee Jan 20 '14 at 05:21

score 2 · Answer 3 · answered Jan 20 '14 at 05:20

2

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.

answered Jan 20 '14 at 05:20

goblin GONE

67,744

$\cos x=0\implies \sin x=?$ does it satisfy the given relation ? – lab bhattacharjee Jan 20 '14 at 05:20
@labbhattacharjee, you're right of course. I've edited. – goblin GONE Jan 20 '14 at 05:22
you mean both $\cos x,\sin x$ are zero, then what will be fate of http://math.stackexchange.com/questions/572585/fundamental-theorem-of-trigonometry ? – lab bhattacharjee Jan 20 '14 at 05:25
@labbhattacharjee, what do you mean? – goblin GONE Jan 20 '14 at 05:26
if both $\cos x,\sin x$ are zero, how $\cos^2x+\sin^2x=1$ will hold ? – lab bhattacharjee Jan 20 '14 at 05:27
@labbhattacharjee, that's not how logic works! What matters is that if you plug an $x$ into line 2 and that same $x$ into line $3$, the resulting truthvalues will agree. Of course, we can simplify 3 by eliminating the vacuously false case, but there's no reason to completely solve the problem. Probably, the OP just needed help getting started. – goblin GONE Jan 20 '14 at 05:29
I understood your point. I thought, it too obvious that both can not be zero :) – lab bhattacharjee Jan 20 '14 at 05:31
@labbhattacharjee, haha yes, it looks kind of odd doesn't it? So be it. – goblin GONE Jan 20 '14 at 05:33

score 1 · Answer 4 · answered Jan 20 '14 at 05:41

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

Trigonometric Equations, Trouble solving.

4 Answers4