Solving $\sqrt3\;\cos(x)= \sin\;(\frac {x}{2})$ I'm currently stuck on this trigonometric equation:
$$\sqrt3 \;\cos(x)= \sin\;(\frac {x}{2})$$
Here's what I've tried so far (2 methods):
Method n.1
$$\sqrt3 \;\cos(x)= \sin\;(\frac {x}{2})$$
$$\sqrt3 \;(1-\sin^2x)= \sin\;(\frac {x}{2})$$
$$\sqrt3-\sqrt3\;\sin^2x= \sin\;(\frac {x}{2})$$
$$2\sqrt3-2\sqrt3\;\sin^2x- \sin\;x=0$$
$$// \; let\; \sin \;(x)=y \;//$$
$$2\sqrt3-2\sqrt3\;y^2- y=0$$
$$y=\frac {1\pm\sqrt{1^2-4(-2\sqrt3)(2\sqrt3)}}{2(-2\sqrt3)}$$
$$y=\frac {1\pm\sqrt{1^2(8\sqrt3)(2\sqrt3)}}{-4\sqrt3}$$
$$y=\frac {1\pm\sqrt{1^2+16\times3}}{-4\sqrt3}$$
$$y=\frac {1\pm\sqrt{49}}{-4\sqrt3}$$
$$y=\frac {1\pm7}{-4\sqrt3}$$
$$y_1=\frac {1+7}{-4\sqrt3}=\frac{-2}{\sqrt3}=\frac{-2\sqrt3}{3}$$
$$y_2=\frac {1-7}{-4\sqrt3}=\frac{-3}{-2\sqrt3}=\frac{\sqrt3}{2}$$
$$\arcsin(\frac{-2\sqrt3}{3})=\nexists$$
$$\arcsin(\frac{\sqrt3}{2})=60°$$
$$x_1= 60°+360°k$$
$$x_1= 120°+360°k$$
Method n.2
$$\sqrt3 \;\cos(x)= \sin\;(\frac {x}{2})$$
$$\sqrt3 \; \cos(x)=\pm \frac{\sqrt{ 1-cos (x)}}{2}$$
Any help is appreciated, thank you in advance! :D
 A: We begin with $\sqrt{3}\cos(x)=\sin(\frac{x}{2})$. We can start by using our half-angle formula to replace $\sin(\frac{x}{2})$ with $\sqrt{\frac{1-\cos(x)}{2}}$. Making the substitution and squaring both sides, we now see that $3\cos^2 (x)=\frac{1-\cos(x)}{2}$. Multiplying both sides by $2$, we get $6\cos^2(x)=1-\cos(x)$, and rearranging the equation, we get $6\cos^2(x)+\cos(x)-1=0$.
Now we can say that $y=\cos(x)$. This turns the equation into $6y^2+y-1=0$. This is an elementary quadratic equation, the solution of which is $y=\frac{-1\pm\sqrt{1^2-4(6)(-1)}}{2(6)}=\frac{-1\pm5}{12}$, or $y=\frac{1}{3}$ and $y=\frac{-1}{2}$.
If $y=\cos(x)$, then of course, $x=\arccos(y)$. And so, $x=\arccos(\frac{1}{3})\approx1.231$ and $x=\arccos(\frac{-1}{2})=\frac{-2\pi}{3}$.
One thing briefly to note is that there's a small amount of strangeness here due to the fact that $\cos(x)=\cos(-x)$. Specifically, a calculator will tell you that $\arccos(\frac{-1}{2})=\frac{2\pi}{3}$, which is true, but does not satisfy our given equation: $\sqrt{3}\cos(\frac{2\pi}{3})=\frac{-\sqrt{3}}{2}$ and $\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}$. In order to remedy this, we switch the sign from positive to negative, keeping the value of the cosine expression negative while switching the sine expression's value to also be negative.
A: hint
If we make the substitution$$x=\pi+2y$$
the equation becomes
$$-\sqrt{3}\cos(2y)=\cos(y)=\sqrt{3}(1-2\cos^2(y))$$
or
$$2\sqrt{3}\cos^2(y)+\cos(y)-\sqrt{3}=0$$
with
$$\Delta=25$$
