If ${\sin }^2\mathrm{x}+\frac{1}{4}{\sin }^{2}3\mathrm{x}=\sin x {\sin }^{2}3\mathrm{x}$ then x=

${\left(\sin \mathrm{x}-\frac{1}{2}{\sin }^{2}3\mathrm{x}\right)}^2+\frac{1}{4}{\sin }^{2}3\mathrm{x}\left(1-{\sin }^{2}3\mathrm{x}\right)=0\text{ or }\sin \mathrm{x}-\frac{1}{2}{\sin }^{2}3\mathrm{x}=0\text{ and }\sin 6\mathrm{x}=0$
From 2sinx=sin²3x and sin6x=0
From here, we choose those values which satisfy the equation 2sinx=sin²3x.
Now, ${\sin }^{2}3\left(\frac{\mathrm{k\pi }}{6}\right)={\sin }^2\frac{\mathrm{k\pi }}{2}=\left\{\begin{array}{l}1,\text{ if }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{k}\text{ is }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ odd }\\ 0,\text{ if }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{k}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ is even }\end{array}\right\}\Rightarrow \sin \mathrm{x}=0\text{ or }\frac{1}{2}\Rightarrow \mathrm{x}=\mathrm{n\pi }\text{ or }\mathrm{x}=\mathrm{n\pi }+\frac{\mathrm{\pi }}{6}(-1{)}^{\mathrm{n}},\mathrm{n}\in \mathrm{z}$