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.
$\endgroup$3 Answers
$\begingroup$$$\implies\sqrt3=\dfrac{\sin x}{\cos x}=?$$
$$\tan x=\tan\dfrac\pi3\implies x=n\pi+\dfrac\pi3$$ where $n$ is any integer.
Choose $n$ such that $x\in[-\pi,\pi]$
$\endgroup$ 1 $\begingroup$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.
If you want to learn more, see here. Hope it helps.
$\endgroup$ $\begingroup$$$\sqrt {3} cosx - sinx = 0$$ Dividing both sides by $2$. $$\frac {\sqrt {3}}{2} cosx - \frac {1}{2} sinx = 0$$ $$sin 60°.cos x- cos 60°.sinx=0$$ $$sin(60-x)°=0$$ $$x=60°,120°$$.
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