Let $f(x)=\frac{5sinx}{4sinx+2cosx}f(x)=\frac{5sinx}{4sinx+2cosx}.$

Then $f^{\text{'}}(x)=f^{\text{'}}(x)=\frac{10}{(4sinx+2cosx{)}^2}.$

The equation of the tangent line to $y=f(x)y=f(x)$ at $a=\frac{}{6}a=\frac{}{6}$ can be written in the form $y=mx+by=mx+b$ where $m=m=$ and

$b=b=$

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