How to represent Hadamard gate as product of Rx and Ry gates?

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As simple as question asks - how to represent Hadamard gate as product of Rx and Ry gates assuming that values for rotation of Rx and Ry gate can be different?

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This is not possible, and one can easily see this via the determinant. All rotation operator gates $$ R_x(\theta)=\begin{pmatrix} \cos\frac\theta2&-i\sin\frac\theta2\\-i\sin\frac\theta2&\cos\frac\theta2 \end{pmatrix}\quad R_y(\theta)=\begin{pmatrix} \cos\frac\theta2&-\sin\frac\theta2\\\sin\frac\theta2&\cos\frac\theta2 \end{pmatrix}\quad R_z(\theta)=\begin{pmatrix} e^{-i\theta/2}&0\\0&e^{i\theta/2} \end{pmatrix} $$ have determinant $1$, regardless of the angle $\theta\in\mathbb R$. The Hadamard gate $$ H=\frac1{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix} $$ however, has determinant $-1$. Now if there existed $k\in\mathbb N$, angles $\theta_1,\ldots,\theta_k\in\mathbb R$ and a gate combination $i_1,\ldots,i_k\in\{x,y,z\}$ such that $H=R_{i_1}(\theta_1)\cdot\ldots\cdot R_{i_k}(\theta_k)$ then $$ -1=\operatorname{det}(H)=\operatorname{det}(R_{i_1}(\theta_1)\cdot\ldots\cdot R_{i_k}(\theta_k))=\operatorname{det}(R_{i_1}(\theta_1))\cdot\ldots\cdot \operatorname{det}(R_{i_k}(\theta_k))=1 $$ because the determinant is multiplicative. But this is a contradiction, hence no decomposition of the Hadamard gate into rotation gates exists.

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We have

$$R_x(\pi)R_y(\pi/2)=\left( \begin{array}{cc} 0 & -i \\ -i & 0 \\ \end{array} \right)\left( \begin{array}{cc} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \end{array} \right)$$

$$=\left( \begin{array}{cc} -\frac{i}{\sqrt{2}} & -\frac{i}{\sqrt{2}} \\ -\frac{i}{\sqrt{2}} & \frac{i}{\sqrt{2}} \\ \end{array} \right)=-i\left( \begin{array}{cc} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \end{array} \right)=-iH$$

This suffices for the purposes of quantum computing since the overall phase of $\frac{3\pi}{2}$ (represented by $-i$) doesn't effect the final measurement output.