Periodic solutions of $x'=x^2-1-\cos t$ Consider $x'=x^2-1-\cos t$. What can be said about the existence of periodic solutions for this equation?
I'm not sure if periodic solutions exist, but if they do, they must have period equal to $ 2\pi$ and $x(0)=x(2k\pi)$ for $\forall k\in\mathbb Z$.
New Edit:
I guess that may use the following lemma:
Lemma: Consider the differential equation $x' = f (t , x)$ where $f(t,   x)$ is continuously differentiable in $t$ and $x$. Suppose that
   $f (t + T, x) = f (t , x)$
    for all t . Suppose there are constants $p$, $q$ such that
    $f (t , p) > 0, f (t , q) < 0$
    for all $t$ then there is a periodic solution $x(t )$ for this equation  with
     $p < x(0) < q$.
Realy, I consider $p=2$ and $q=0$ but $f(t,q)=-1-\cos t\leq 0$ and this inequality is not strictly. 

Hint:
As explained in a now deleted answer, the equation can be linearized with
$$x(t)=-\frac{y'(t)}{y(t)}$$
giving
$$y''(t)=(1+\cos(t))y(t).$$
Thanks to linearity and homogeneity, the general solution will be of the form
$$y(t)=C_0f(t)+C_1g(t),$$ and 
$$x(t)=-\frac{C_0f'(t)+C_1g'(t)}{C_0f(t)+C_1g(t)}=-\frac{f'(t)+Cg'(t)}{f(t)+Cg(t)}.$$
Then we can achieve $x(2\pi)=x(0)$ and form a periodic function by solving
$$\frac{f'(0)+Cg'(0)}{f(0)+Cg(0)}=\frac{f'(2\pi)+Cg'(2\pi)}{f(2\pi)+Cg(2\pi)},$$ a quadratic equation in $C$.
$$f'(0)f(2\pi)-f'(2\pi)f(0)+((g'(0)f(2\pi)-f'(2\pi)g(0)+f'(0)g(2\pi)-g'(2\pi)f(0))C+(g'(0)g(2\pi)-g'(2\pi)g(0))C^2=0.$$
