Recover an inverse-cosine tangent from its Maclaurin coefficients

$\cos^2y=1-e^x/4$ gives $-2\cos y\sin y\,y^{\prime}=-e^x/4$, hence $\sin(2y)y^{\prime}=e^x/4$. The square root is present in the original printed question, though absent from its native extraction.

A further derivative gives $\sin(2y)y^{\prime\prime}+2\cos(2y)(y^{\prime})^2=e^x/4$. The branch condition gives $y(0)=\pi/6$, $y^{\prime}(0)=1/(2\sqrt3)$ and $y^{\prime\prime}(0)=1/(3\sqrt3)$. Thus

$\begin{aligned}&y\\ & =\frac\pi6+\frac{x}{2\sqrt3}+\frac{x^2}{6\sqrt3}+O(x^3).\end{aligned}$

The tangent has point $(0,\pi/6)$ and slope $1/(2\sqrt3)$, so $y-\pi/6=x/(2\sqrt3)$, or $y=x/(2\sqrt3)+\pi/6$. This is the exact tangent, not a quadratic approximation to it.