Is there any method to solve the equation above for $x$?
$$\alpha_1\sin{(x})+\beta_1\cos{(x)}+\alpha_2\sin{(2x)}+\beta_2\cos{(2x)}=c$$
This would be a trivial problem if it was only the first pair or second pair of functions. Perhaps it's also a trivial question as it stands and I'm simply unaware. All I know is that WolframAlpha was really chugging and couldn't give a clean answer before timing out.
Recall the double-angle formulas $\cos(2x) = \cos^2(x) - \sin^2(x)$ and $\sin(2x) = 2 \cos(x) \sin(x)$, so your equation becomes:
$$\alpha_1 \sin(x) + \beta_1 \cos(x) + \alpha_2 (\cos^2(x) - \sin^2(x)) + 2\beta_2 \cos(x) \sin(x) = c$$
For the sake of brevity, let $k = \cos(x)$ and $s = \sin(x)$.
$$\alpha_1 s + \beta_1 k + \alpha_2 (k^2 - s^2) + 2\beta_2 k s = c$$
But $k^2 + s^2 = 1$, so $k^2 = 1 - s^2$.
$$\alpha_1 s + \beta_1 k + \alpha_2 (1 - 2s^2) + 2\beta_2 k s = c$$
Solving for $k$ gives:
$$k = \frac{c - \alpha_1 s - \alpha_2 + 2 \alpha_2 s^2}{\beta_1 + 2\beta_2 s}$$
Squaring both sides gives:
$$k^2 = 1 - s^2 = \frac{(c - \alpha_1 s - \alpha_2 + 2 \alpha_2 s^2)^2}{(\beta_1 + 2\beta_2 s)^2}$$
which, after a bunch of algebra (that I'm not 100% sure I did correctly), expands to a quartic equation in terms of $s$.
$$-4(\alpha_2^2 + \beta_2^2)s^4 + 4 \alpha_1 \alpha_2 s^3 + (4 - 4 \alpha_2 c - \alpha_1^2 + \alpha_2^2 + 4 \beta^2 - \beta_1^2 - 4 \beta_1 \beta_2 )s^2 + (2 \alpha_1 c - 2 \alpha_1 \alpha_2)s + (\beta_1^2 + 4 \beta_1 \beta_2 + 2\alpha_2 c - \alpha_2^2 - c^2) = 0$$
From here, it's “just” a matter of plugging the coefficients into the Quartic Formula to solve for $s$, and then taking the inverse sine of that to get $x$.
Fortunately, you didn't ask what the solutions were, just “is there a way to solve” your equation. And since quartic polynomials (unlike quintic and higher) have “closed-form” solutions, the answer is yes :-)
