Answer
$$
y=\sqrt{1+4 \sin x}, \,\,\,\, (0,1)
$$
The equation of the tangent line is
$$ y=2 x+1.$$
The equation of the normal line is
$$ y=-\frac{1}{2} x+1$$
Work Step by Step
$$
y=\sqrt{1+4 \sin x}, \,\,\,\, (0,1)
$$
we have
$$
\begin{aligned} y^{\prime}&=\frac{1}{2}(1+4 \sin x)^{-1 / 2} \cdot 4 \cos x \\
&=\frac{2 \cos x}{\sqrt{1+4 \sin x}}
\end{aligned}
$$
So the slope of the tangent line at $(0,1)$ is
$$
\left.\frac{d y}{d x}\right|_{x=0}=\frac{2 \cos (0)}{\sqrt{1+4 \sin (0)}}=\frac{2}{\sqrt{1}}=2
$$
so an equation of the tangent line is
$$y-1=2(x-0), \,\,\ \text{or} \,\,\, y=2 x+1.$$
The slope of the normal line is $-\frac{1}{2},$
so an equation of the normal line is
$$y-1=-\frac{1}{2}(x-0), \,\,\,\ \text{ or} \,\,\, y=-\frac{1}{2} x+1$$