= Solution
Recognize the left-hand side as a <product rule> <derivative>, and use $\tfrac12\sin2x=\sin x\cos x$:
$$
\frac{d}{dx}(y\sin x)=\sin x\cos x,\qquad y\sin x=\frac12\sin^2x+C.
$$
At $x=\pi/6$, both $y\sin x$ and $\tfrac12\sin^2x$ equal $1/8$, so $C=0$. Therefore
$$
\boxed{y(x)=\frac12\sin x.}
$$
Dividing by $\sin x$ first obtains this solution on $(0,\pi)$, the interval containing the <initial condition>. The explicit expression is smooth on the entire real line and satisfies the original, undivided <differential equation> even where $\sin x=0$.
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