A spin-$1/2$ particle is subject to an external magnetic field $$\mathbf{B}\left(t\right)=B\left(\mathbf{i}\cos{\omega t}-\mathbf{j}\sin{\omega t}\right) + B_0\mathbf{k} \; \left(B,B_0\in\mathbb{R^+}\right).$$ The state ket obeys the equation $$i\hbar\frac{\rm{d}}{\rm{d}t}\left|\alpha,t\right\rangle=-\gamma\mathbf{S\cdot B}\left|\alpha,t\right\rangle, \; \left(\gamma\in\mathbb{R^+}\right)$$ with initial condition $$\left|\alpha,0\right\rangle=\left|\frac{1}{2}\frac{1}{2}\right\rangle\equiv\left|+\right\rangle$$ in the $\left\{\mathbf{S}^2,S_z\right\}$ basis. Calculate $\left|\alpha,t\right\rangle$.
Writing the Hamiltonian as $$H\left(t\right)=-\gamma\mathbf{S\cdot B}\equiv H_0 + V\left(t\right),$$ I tried to use interaction picture, $$\left|\alpha,t\right\rangle_I\equiv e^{i H_0 t/\hbar}\left|\alpha,t\right\rangle.$$Now the ket obeys the differential equation $$i\hbar\frac{\mathrm{d}}{\mathrm{d}t}\left|\alpha,t\right\rangle_I=V_I\left(t\right)\left|\alpha,t\right\rangle_I,$$ with initial condition $$\left|\alpha,0\right\rangle_I =\left|\alpha,0\right\rangle=\left|+\right\rangle$$ and $$V_I\left(t\right)\equiv e^{iH_0 t/\hbar}V\left(t\right)e^{-iH_0 t/\hbar}.$$ So, since $$H_0\equiv -\gamma B_0 S_z=-\omega S_z, \;\; V\left(t\right)\equiv-\gamma B \left(S_x \cos{\omega t} - S_y \sin{\omega t}\right),$$ I got the coupled system of ODEs $$i\hbar\frac{\mathrm{d}}{\mathrm{d}t}\left|\alpha,t\right\rangle_I=-\frac{\gamma B\hbar}{2}\begin{pmatrix}0&1\\1&0\end{pmatrix}\left|\alpha,t\right\rangle_I$$ for some ket of the form $$\left|\alpha,t\right\rangle_I=\begin{pmatrix}a\left(t\right)\\b\left(t\right)\end{pmatrix}.$$ Differentiating again with respect to time, I got $$\frac{\mathrm{d}^2}{\mathrm{d}t^2}a\left(t\right)+\omega_1^2 a\left(t\right)=0,$$ whose solution, taking initial condition into account, is $$a\left(t\right)=a_0 \cos{\omega t}.$$ So, $$b\left(t\right)=\frac{2i\omega a_0}{\gamma B}\sin{\omega t.}$$ To deal with a normalized state ket, one must have $$1=\left|a\left(t\right)\right|^2 + \left|b\left(t\right)\right|^2=\left|a_0\right|^2\left(\cos^2{\omega t}+\left(\frac{2\omega}{\gamma B}\right)^2\sin^2{\omega t}\right),$$ which, because $\omega=\gamma B_0$, brings the conditions $$\left|a_0\right|^2=1, \;\; B=2 B_0.$$ But I find very strange the second condition. Is there any problem with my idea or with my calculations?