Find angle $\alpha$ in equation containing weighted sum of $\sin(\alpha)$ and $\cos(\alpha)$. The context of this question comes from calculating the Fidelity drop of quantum systems after applying a rotation with some arbitrary angle $\alpha$ on the state vector of the system, see equation below:
$$\begin{eqnarray}
\Delta F = &sin(\alpha)&(\frac{(a_{1} + a_{4})(b_{2} + b_{3}) - (a_{2} + a_{3})(b_{1} + b_{4})}{2}) \nonumber \\
&-&cos(\alpha)(\frac{(b_{2} + b_{3})^{2} + (a_{2} + a_{3})^{2} - F}{2}) \nonumber \\
&+&(\frac{(b_{2} + b_{3})^{2} + (a_{2} + a_{3})^{2} - F}{2}) \nonumber
\end{eqnarray}$$
$\Delta F$, $F$, $a_i$ and  $b_j$, $1\leq i,j\leq 4$ are known.
Could someone explain to me how I could get the angle $\alpha$, if that is even possible?
I will be very grateful for your help!
I have another similar problem, but with a weighted sum of $(cos(\alpha))^{2}$,  $(sin(\alpha))^{2}$, and $cos(\alpha)sin(\alpha)$ in which I could also get some help.

Let us have the general (and prettier) equation $$B=m\sin\alpha -n\cos \alpha+K.$$
Transpose $K$ and divide the whole equation by $\sqrt{m^2+n^2}$:$$\frac{B-K}{\sqrt{m^2+n^2}}=\frac{m}{\sqrt{m^2+n^2} }\sin\alpha-\frac{n}{\sqrt{m^2+n^2} }\cos\alpha$$
Now, $\Bigg |\dfrac {m}{\sqrt{m^2+n^2}}\Bigg |<1$ and so there must exist some angle A such that $\cos A= \dfrac {m}{\sqrt{m^2+n^2}}$ and also $\sin A= \dfrac {n}{\sqrt{m^2+n^2}}$ and $\tan A=\dfrac nm$. We have now,
$$\frac{B-K}{\sqrt{m^2+n^2}}=\sin\alpha\cos A-\cos\alpha\sin A=\sin (\alpha-A).$$ Thus, $$\alpha=A+\arcsin\left(\frac{B-K}{\sqrt{m^2+n^2}}\right)$$$$=\arctan\frac nm+ \arcsin\left(\frac{B-K}{\sqrt{m^2+n^2}}\right).$$
EDIT: Comparing the general equation to the OP’s equation, we have $K=n$.
