<s>[INST] Trigonometric system with different coefficients I have a trigonometric system as below and I want to solve it for $x$.
\begin{gather}
A\cos(x)+B\cos(2x)+C\sin(2x)  = D \\
-C\cos(2x)-A\sin(x)+B\sin(2x) = F
\end{gather}
Can anyone suggest a solution?
Regards.
 [/INST] Note $(1)$ the first equation and $(2)$ the second. Multiplying $(1)$ by $C$ and $(2)$ by $B$ the sum of two resulting equations give $$A\cos(\theta-x)+\cos(2x)=\frac{BD-CF}{B^2+C^2}=M$$ Similarly you get $$A\sin(\theta-x)+\sin(2x)=\frac{CD-BF}{B^2+C^2}=N$$ where $\sin(\theta)=\dfrac CB$. Now taking the square and adding you get
$$A^2(1)=(M-\cos(2x))^2+(N-\sin(2x))^2$$ so you have
$$M\cos(2x)+N\sin(2x)=\frac{M^2+N^2+1-A^2}{2}$$ which is a standard equation you can solve. In fact you do have
$$\sin(\phi+x)=\frac{M^2+N^2+1-A^2}{2(M^2+N^2)}$$ where $\sin(\phi)=\dfrac{M}{M^2+N^2}$.
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