Why am I getting a 3rd answer for $\sin(x) + \sqrt{3}\cos(x) = 1$?

Question

I'm getting an extra solution to $sin(x) + \sqrt{3}cos(x)$ = 1 (which is $\frac{7\pi}{6}$) that doesn't show up in the graph or is wrong. Where am I going wrong with my thinking? I'm also trying to understand this intuitively.

So I start with $sin(x) + \sqrt{3}cos(x) = 1$.

I know that I can define $\sin(x)$ in terms of $\cos(x)$. So I get $\sin(x) = 1 - \sqrt{3}\cos(x)$.
I also know that I can get a true value for $\cos(x)$ by using the Pythagorean theorem with all terms defined in some form of $\cos(x)$. So we have $\cos^2(x)+(1-\sqrt{3}\cos(x))^2=1$

which becomes $4\cos^2(x)-2\sqrt{3}\cos(x) = 0$

Now I factor out $2\cos(x)$ and get $2\cos(x)(2\cos(x)-\sqrt{3}) = 0$
We then have two values for $\cos(x)$ which are $\cos(x) = 0$ and $\cos(x) = \frac{\sqrt{3}}{2}$

We plug these values back into our original equation of $\sin(x) + \sqrt{3}\cos(x)$ and get the following:

$\sin(x) + \sqrt{3}(0)=1$ and $\sin(x) + \sqrt{3}(\frac{\sqrt{3}}{2})=1$
This becomes $\sin(x) = 1$ and $\sin(x) = \frac{-1}{2}$.

When $\sin(x) = 1$, $x = \frac{\pi}{2}$ and when $\sin(x) = \frac{-1}{2}$, $x = \frac{7\pi}{6} + 2\pi k$ (which does not show up on the graph) and $x = \frac{11\pi}{6} + 2\pi k$ which does show up on the graph. Why am I getting this extra answer?

When you found $\sin x=-1/2$, you had already assumed $\cos x=\sqrt3/2$. — Empy2, Nov 09 '20 at 13:41
Actually, $\frac{7\pi}6$ is not a solution of the equation $\sin(x)+\sqrt3\cos(x)=1$, since $\sin\left(\frac{7\pi}6\right)+\sqrt3\cos\left(\frac{7\pi}6\right)=-2$. — José Carlos Santos, Nov 09 '20 at 13:46

score 4 · Accepted Answer · answered Nov 09 '20 at 13:38

4

Hint:

By squaring, you are introducing

$$\sin x=-(1-\sqrt3\cos x)$$ as well

See When do we get extraneous roots?

So, avoid squaring whenever possible

For this case,

we can use Weierstrass substitution
or Double angle formula, $$\sqrt3\left(\cos^2\dfrac x2-\sin^2\dfrac x2\right)=\left(\cos\dfrac x2-\sin\dfrac x2\right)^2$$
use Solving trigonometric equations of the form $a\sin x + b\cos x = c$

answered Nov 09 '20 at 13:38

lab bhattacharjee

274,582

That wasn't the problem here, as it would lead to $\sin x=1/2$, or $x=\pi/6$ – Empy2 Nov 09 '20 at 13:45
I think at a high level you are spot on. It looks more like a problem with finding a solution for cos(x) first and plugging it back into the original problem. Since there will always be two angles for cos(x) unless cos(x) = -1 or 1. I've just never run into this before. It just seems strange to me that this would happen. I found an example in my book converting the original equation into a Sinusoid. So maybe I will just go with that. I still don't understand fully why my original approach failed but I guess it is what it is. – maybedave Nov 09 '20 at 19:28

score 1 · Answer 2 · answered Nov 09 '20 at 14:15

You found the solutions

$$\cos x=0,\cos x=\frac{\sqrt 3}2$$ and $$\sin x=1,\sin x=-\frac12,$$ which might make you think that there are in fact five of them ($\frac\pi2,\frac{3\pi}2,\frac\pi6,\frac{7\pi}6,\frac{11\pi}6$). In other cases, you would even find six solutions this way !

But you forget that these hold in pairs, giving only two possible angles.

score 0 · Answer 3 · answered Nov 09 '20 at 13:56

You have : $$\sin x+\sqrt{3}\cos x=1\\\frac{1}{2}\sin x+\frac{\sqrt{3}}{2}\cos x=\frac{1}{2}\\ \sin{\frac{\pi}{6}}\sin x+\cos{\frac{\pi}{6}}\cos x=\frac{1}{2} \\ \cos\left(x-\frac{\pi}{6}\right)=\cos{\frac{\pi}{3}} $$

So you get : $$x-\frac{\pi}{6}= \frac{\pi}{3},\text{or }x-\frac{\pi}{6}=-\frac{\pi}{3}\\ x=\frac{\pi}{2}+2k\pi \text{ or : }x=\frac{11\pi}{6}+2k\pi$$

Why am I getting a 3rd answer for $\sin(x) + \sqrt{3}\cos(x) = 1$?

3 Answers3