<s>[INST] Solving a nonlinear periodic ODE Is it possible to solve the below ODE for arbitrary real values of $c$?
$$2 (\frac{d \theta}{dx})^2 [\sin(2\theta) -c \cos(2 \theta)]-[\cos(2\theta) +c \sin(2 \theta)]\frac{d^2}{dx^2}\theta=0,$$
where $\theta=\theta(x)$. Also, $\theta$ itself does not have meaning, but ($\cos(2\theta),\sin(2\theta)$) are important and need to be continuous. There is no boundary condition since $x$ is not constrained by a boundary but goes to infinity.  It means that the solution must cover the range $(-\infty,\infty)$.
 [/INST] The given equation can be easily integrated:
$$-\dfrac12\left((\cos 2\theta+c\sin2\theta)\theta'\right)'=0\Rightarrow
(\cos 2\theta+c\sin2\theta)\theta' = \dfrac12C_1,\\
(\cos 2\theta+c\sin2\theta)\,\mathrm d(2\theta) = C_1\,\mathrm dx\Rightarrow 
\sin 2\theta-c\cos2\theta = C_1 x+C_2 = y,$$
wherein the constants $C_1,C_2\ $ can be defined from the initial conditions.
Let
$$t = \tan\theta\Rightarrow \cos2\theta = \dfrac{\cos^2\theta-\sin^2\theta}{\cos^2\theta+\sin^2\theta} = \dfrac{1-t^2}{1+t^2},\quad \sin2\theta = \dfrac{2\sin\theta\cos^2\theta}{\cos^2\theta+\sin^2\theta} = \dfrac{2t}{1+t^2},$$
$$2t-c(1-t^2) = (1+t^2)y,\quad (c-y)t^2 + 2t - (c+y) = 0,$$
with the common solutions
$$\small t_{1,2} = \dfrac{1\pm\sqrt{1+c^2-y^2}}{c-y},$$
$$\cos2\theta = \dfrac{-cy\mp\sqrt{1+c^2-y^2}}{c^2+1},\ \sin2\theta = \dfrac{-y\pm c\sqrt{1+c^2-y^2}}{c^2+1}, \tag1$$
and the special case
$$y=c\Rightarrow t=c,\quad \cos2\theta =\dfrac{1-c^2}{1+c^2},\quad \sin2\theta =\dfrac{2c}{1+c^2}.\tag2$$
If $y^2 > 1+c^2,\ $ then the solutions $(1)$ are complex. This fact points to the bounded values of the variable $x.$
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