= Solution
Recover $L_1$ from the given <angular momentum raising operator> and <angular momentum lowering operator>:
$$
L_1=\frac12(L_++L_-)=i\hbar\left(\sin\phi\,\partial_\theta+\cot\theta\cos\phi\,\partial_\phi\right).
$$
This is the generator of <rotations> about the first coordinate axis. A natural candidate is the angular function pointing along that axis, $f_x=\sin\theta\cos\phi$. Direct substitution gives
$$
L_1f_x=i\hbar\left(\sin\phi\cos\theta\cos\phi-\cot\theta\cos\phi\sin\theta\sin\phi\right)=0.
$$
Also $f_x=(f_{+1}+f_{-1})/2$, so <linearity> and part (ii) give $L^2f_x=2\hbar^2f_x$. Thus an answer is $\sin\theta\cos\phi$. Its squared <norm> on the sphere is
$$
\int_0^{2\pi}\cos^2\phi\,d\phi\int_0^\pi\sin^3\theta\,d\theta=\pi\frac43=\frac{4\pi}3,
$$
so the normalized <quantum state> is
$$
\boxed{\psi_x(\theta,\phi)=\sqrt{\frac3{4\pi}}\sin\theta\cos\phi,\qquad L^2\psi_x=2\hbar^2\psi_x,\qquad L_1\psi_x=0.}
$$
The supplied starting state gives an alternative derivation with a changed <quantization axis>. Put $g=\cos\theta+i\sin\theta\sin\phi$ and $K_+=L_2+iL_3$. The <angular momentum commutation relations> give $[L_1,K_+]=\hbar K_+$ and $[L^2,K_+]=0$, so $K_+$ raises the $L_1$ projection from $-\hbar$ to zero. Direct differentiation gives
$$
L_2g=i\hbar\sin\theta\cos\phi,\qquad L_3g=\hbar\sin\theta\cos\phi,\qquad K_+g=2i\hbar\sin\theta\cos\phi,
$$
which is nonzero and proportional to the same $\psi_x$. This also connects the construction directly to the given state.
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