# Equating trigonometric identities: $\sqrt3\cos x - \sin x=0$

So I was given the equation below and was told to solve for x.

$$\sqrt3\cos(x) - \sin(x) = 0,x \in [-\pi, \pi]$$

Naturally I rearranged the equation to be:

$$\sqrt3\cos(x) = \sin(x)$$

Once I had it in this form, it was obvious to me that this equation would be true for when $\cos(x) = \frac{1}{2}$ and $\sin(x) = \frac{\sqrt3}{2}$ or the negative version of each of these. I then proceeded to solve the question.

However, I feel that this step is "not good enough" in terms of a mathematical solution. Is there a better way that we can find $x$ without having to just mentally guess what it is going to be? Obviously it is fairly easy to do in this case, but I don't want to be blindsided by a harder question of the same type in the future.

Any more formal proofs/solutions would be greatly appreciated.

$$\implies\sqrt3=\dfrac{\sin x}{\cos x}=?$$

$$\tan x=\tan\dfrac\pi3\implies x=n\pi+\dfrac\pi3$$ where $n$ is any integer.

Choose $n$ such that $x\in[-\pi,\pi]$

• Awesome, didn't realize it would be that simple. Thanks! – Benjamin Lowry Jan 11 '17 at 13:27

We generally solve as follows: $$\sqrt {3}\cos x =\sin x$$ $$\tan x =\sqrt {3} =\tan 60^\circ$$ The general solution for this equation is $x =n\pi +\frac {\pi}{3}, n\in \mathbb Z$. Substituting for the values of $n$, we get solutions in the required range.

$$\sqrt {3} cosx - sinx = 0$$ Dividing both sides by $2$. $$\frac {\sqrt {3}}{2} cosx - \frac {1}{2} sinx = 0$$ $$sin 60°.cos x- cos 60°.sinx=0$$ $$sin(60-x)°=0$$ $$x=60°,120°$$.