Problem - 3655
Find the function $f(x)$ such that $f(0)=1$, $f(\frac{\pi}{2})=2$, and for any $x, y\in\mathbb{R}$, $$f(x+y)+f(x-y)=2f(x)\cos y$$
Setting $x=0, y=\alpha$:
$$f(\alpha) + f(-\alpha)=2f(0)\cos\alpha = 2\cos\alpha\qquad (1)$$
Setting $x=\frac{\pi}{2}+\alpha, y=\frac{\pi}{2}$:
$$f(\pi + \alpha) + f(\alpha)=0\qquad (2)$$
Setting $x=\frac{\pi}{2}, y=\frac{\pi}{2}+\alpha$:
$$f(\pi + \alpha) + f(-\alpha)=-2f(\frac{\pi}{2})\sin\alpha = -4\sin\alpha\qquad (3)$$
Now, $(1) + (2) - (3)$ yields:
$$2f(\alpha) = 2\cos\alpha + 4\sin\alpha\implies f(\alpha) = \cos\alpha + 2\sin\alpha$$
$$\therefore\quad f(x) = \boxed{\cos x + 2\sin x}$$