Solve $2+ \cos{\frac{3x}{2}} + \sqrt{3} \sin{\frac{3x}{2}} = 4\sin^2{\frac{x}{4}}$ $$2+ \cos{\frac{3x}{2}} + \sqrt{3} \sin{\frac{3x}{2}} = 4\sin^2{\frac{x}{4}}$$
My try:
$$ \cos{\frac{3x}{2}} + \sqrt{3} \sin{\frac{3x}{2}} = \sqrt{4}\left(\frac{\sqrt{3}}{2}  \sin{\frac{3x}{2}} + \frac{1}{2}\cos{\frac{3x}{2}}\right) = 2\sin\left({\frac{3x}{2} + \frac{\pi}{6}}\right)
$$
$$
2+  2\sin\left({\frac{3x}{2} + \frac{\pi}{6}}\right) = 4\sin^2{\frac{x}{4}}
$$
$$
2\sin\left({\frac{3x}{2} + \frac{\pi}{6}}\right) + \cos{\frac{x}{2}}=0
$$
And then...
 A: We have $$2+ \cos{\frac{3x}{2}} + \sqrt{3} \sin{\frac{3x}{2}} = 4\sin^2{\frac{x}{4}}$$ $$1+\frac{1}{2}\cos{\frac{3x}{2}} + \frac{\sqrt{3}}{2} \sin{\frac{3x}{2}} = \frac{4}{2}\sin^2{\frac{x}{4}}$$
$$\cos{\frac{3x}{2}}\cos \frac{\pi}{3} +\sin{\frac{3x}{2}}\sin\frac{\pi}{3} =-\left(1-2\sin^2{\frac{x}{4}}\right)$$
$$\cos\left(\frac{3x}{2}-\frac{\pi}{3}\right)=-\cos \frac{x}{2}=\cos \left(\pi-\frac{x}{2}\right)$$ Now, writing the general solution as follows $$\frac{3x}{2}-\frac{\pi}{3}=2n\pi\pm \left(\pi-\frac{x}{2}\right)$$ taking positive sign, we get $$\frac{3x}{2}-\frac{\pi}{3}=2n\pi+ \left(\pi-\frac{x}{2}\right)$$  $$2x=2n\pi+ \frac{4\pi}{3}$$  $$\color{blue}{ x=n\pi+\frac{2\pi}{3}}$$ taking negative sign, we get
$$\frac{3x}{2}-\frac{\pi}{3}=2n\pi- \left(\pi-\frac{x}{2}\right)$$
$$x=2n\pi-\pi+\frac{\pi}{3}$$
$$\color{blue}{ x=2n\pi-\frac{2\pi}{3}}$$ Where, $n$ is any integer.
A: $2+2\sin(\frac{3x}{2}+\frac{\pi}{6})=4 \sin^2(\frac{x}{4})$ simplifies to $2\sin(\frac{3x}{2}+\frac{\pi}{6})+2 cos(\frac{x}{2})=0$
$\sin(\frac{3x}{2}+\frac{\pi}{6})=-cos(\frac{x}{2})$
$\sin(\frac{3x}{2}+\frac{\pi}{6})=sin(\frac{3\pi}{2}-\frac{x}{2})$
I hope you can solve further.
