Solving $\sqrt3\;\cos(x)= \sin\;(\frac {x}{2})$

algebra-precalculustrigonometry

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! 😀

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.