Quantum Computing 2.2 -- What is my quantum gate?

Matrix multiplication yields $\begin{aligned} H \cdot R_\phi \cdot H &= \frac{1}{2} \left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right) \cdot \left( \begin{array}{cc} 1 & 0 \\ 0 & e^{i \phi} \end{array} \right) \cdot \left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right) \\ &= \frac{1}{2} \left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right) \cdot \left( \begin{array}{cc} 1 & 1 \\ e^{i \phi} & -e^{i \phi} \end{array} \right) \\ &= \frac{1}{2} \left( \begin{array}{cc} 1 + e^{i \phi} & 1 - e^{-i \phi} \\ 1 - e^{-i \phi} & 1 + e^{i \phi}\end{array} \right) \\ &= \frac{e^{i \phi/2}}{2} \left( \begin{array}{cc} e^{-i \phi/2} + e^{i \phi/2} & e^{-i \phi/2} - e^{i \phi/2} \\ e^{-i \phi/2} - e^{i \phi/2} & e^{-i \phi/2} + e^{i \phi/2} \end{array} \right) \\ &= e^{i \phi/2} \left( \begin{array}{cc} \cos \frac{\phi}{2} & -i \sin \frac{\phi}{2} \\ -i \sin \frac{\phi}{2} & \cos \frac{\phi}{2} \end{array} \right) \end{aligned}$ Therefore, the final state is $H \cdot R_\phi \cdot H |0\rangle = e^{i \phi/2} \cdot \left( \cos \frac{\phi}{2} |0\rangle -i \sin \frac{\phi}{2} |1 \rangle \right)$ The probability for measuring $|0\rangle$ results to $\langle 0| H \cdot R_\phi \cdot H |0\rangle = \cos^2 \frac{\phi}{2} \stackrel{!}{=} \frac{3}{4} \quad \Rightarrow \quad \boxed{\phi = \dfrac{\pi}{3}}$