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

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
$$\Delta=25$$
