$(\frac{\pi}{2},-\sin^2 1)$ and parallel to x axis.
I solved the equation and it basically gets reduced to
$$y=-\sin^2 1$$ It’s clearly parallel to the x axis and passes through $-\sin^2 1$, but where does $\frac{\pi}{2}$ come from?
I know a similar problem has already been asked on this site, but it doesn’t solve my particular query
Use the fact that $\cos a\cos b=\dfrac{1}{2}(\cos(a+b)+\cos(a-b)).$ Doing so yields $\begin{align}y&= \dfrac{1}{2}(\cos(2x+2)+\cos2)-\dfrac{1}{2}(\cos(2x+2)+\cos0)\\ &=\dfrac{1}{2}(1-2\sin^2 1 - 1)\\ &=-\sin^2 1.\end{align}$
Hence it passes through the desired point $(\frac\pi2, -\sin^2 1)$