I'll tell you how to create any two qubit pure state you might ever be interested in. Hopefully you can use it to generate the state you want.
Using a single qubit rotation followed by a cnot, it is possible to create states of the form
$$ \alpha \, |0\rangle \otimes |0\rangle + \beta \, |1\rangle \otimes |1\rangle .$$
Then you can apply an arbitrary unitary, $U$, to the first qubit. This rotates the $|0\rangle$ and $|1\rangle$ states to new states that we'll call $|a_0\rangle$ and $|a_1\rangle$,
$$U |0\rangle = |a_0\rangle, \,\,\, U |1\rangle = |a_1\rangle$$
Our entangled state is then
$$\alpha \, |a_0\rangle \otimes |0\rangle + \beta \, |a_1\rangle \otimes |1\rangle .$$
We can similarly apply a unitary to the second qubit.
$$V |0\rangle = |b_0\rangle, \,\,\, V |1\rangle = |b_1\rangle$$
which gives us the state
$$\alpha \, |a_0\rangle \otimes |b_0\rangle + \beta \, |a_1\rangle \otimes |b_1\rangle .$$
Due to the Schmidt decomposition, it is possible to express any pure state of two qubits in the form above. This means that any pure state of two qubits, including the one you want, can be created by this procedure. You just need to find the right rotation around the x axis, and the right unitaries $U$ and $V$.
To find these, you first need to get the reduced density matrix for each of your two qubits. The eigenstates for the density matrix of your first qubit will be your $|a_0\rangle$ and $|a_1\rangle$. The eigenstates for the second qubit will be $|b_0\rangle$ and $|b_1\rangle$. You'll also find that $|a_0\rangle$ and $|b_0\rangle$ will have the same eigenvalue, which is $\alpha^2$. The coefficient $\beta$ can be similarly derived from the eigenvalues of $|a_1\rangle$ and $|b_1\rangle$.