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

${\sin }^2\mathrm{x}+\frac{1}{4}{\sin }^{2}3\mathrm{x}=\sin {\mathrm{xsin}}^{2}3\mathrm{x}$
${\sin }^2\mathrm{x}+\frac{1}{4}{\sin }^{4}3\mathrm{x}-\sin {\mathrm{xsin}}^{2}3\mathrm{x}+\frac{1}{4} {\sin }^{2}3\mathrm{x}-\frac{1}{4}{\sin }^{4}3\mathrm{x}-=0$
$$\begin{gathered} {\left(\sin \mathrm{x}-\frac{1}{2}{\sin }^{2}3\mathrm{x}\right)}^2+\frac{1}{4}{\sin }^{2}3x {\cos }^{2}3x=0 \\ {\left(\sin \mathrm{x}-\frac{1}{2}{\sin }^{2}3\mathrm{x}\right)}^2+\frac{1}{16}{\sin }^{2}6x=0 \end{gathered}$$
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, ${\sin }^{2}3\left(\frac{\mathrm{k\pi }}{6}\right)={\sin }^2\frac{\mathrm{k\pi }}{2}=\left\{\begin{array}{c}1,\text{ if }\mathrm{k}\text{ is odd }\\ 0,\text{ if }\mathrm{k}\text{ is even }\end{array}\right\}$
$\begin{array}{l}\Rightarrow \sin \mathrm{x}=0\text{ or }\frac{1}{2}\\ \Rightarrow \mathrm{x}=\mathrm{n\pi }\text{ or }\mathrm{x}=\mathrm{n\pi }+\frac{\mathrm{\pi }}{6}(-1{)}^{\mathrm{n}},\mathrm{n}\in \mathrm{Z}\end{array}$