Given $f''(x) = -4f(x)$ for all $x \in$ $\mathbb{R}$ and $f(0) = f'(0) = 1$, how do I find $f(x)$ Let $f$ be a function so that $f''(x) = -4f(x)$ for all $x \in \mathbb{R}$ and $f(0) = f'(0) = 1$.
i) Prove that $2f(x)\sin(2x)+f'(x)\cos(2x)=1$ for all $x \in \mathbb{R}$
ii) Prove that $2f(x)\cos(2x)-f'(x)\sin(2x)=2$ for all $x \in \mathbb{R}$
iii) Using (i) and (ii), prove that $f(x)=\frac12\sin(2x)+\cos(2x)$ for all $x \in \mathbb{R}$
 A:  (This part is obsolete after the question has been fixed)
The exercise is wrongly stated. Indeed, if the identities 1 and 2 hold, we immediately get $f(x)\sin 2x$ is constant, which doesn't agree with 3.
The identity 2 should read $2f(x)\cos2x-f'(x)\sin2x=2$

Let $g(x)=2f(x)\sin2x+f'(x)\cos2x$. Then
$$
g'(x)=2f'(x)\sin2x+4f(x)\cos2x+f''(x)\cos2x-2f'(x)\sin2x=0
$$
Since $g(0)=1$, you're done.
The second function is
$$
h(x)=2f(x)\cos2x-f'(x)\sin2x
$$
and by differentiating we obtain
$$
h'(x)=2f'(x)\cos2x-4f(x)\sin2x-f''(x)\sin2x-2f'(x)\cos2x=0
$$
and $h(0)=2$.
Now consider
$$
\begin{cases}
2f(x)\sin2x+f'(x)\cos2x=1 \\[4px]
2f(x)\cos2x-f'(x)\sin2x=2
\end{cases}
$$
Multiply the first relation by $\sin2x$ and the second relation by $\cos2x$ and sum.
A: Hint for i) and ii), compute the derivative of $2f(x)sin(2x)+f'(x)cos(2x)$ and $-2f(x)sin(2x)+f'(x)cos(2x)$ and shows it is $0$. 
$g(x)=2f(x)sin(2x)+f'(x)cos(2x)$, $g'(x)=2f'(x)sin(2x)+4f(x)cos(2x)+f"(x)cos(2x)-2f'(x)sin(2x)=0$ since $f"=-4f$. So it is constant, $g(0)=1$.
Now solve the linear system:
$2f(x)sin(2x)+f'(x)cos(2x)=1$
$-2f(x)sin(2x)+f'(x)cos(2x)=1$
whose variables are $f(x)$ and$f'(x)$.
A: 
you can solve the 1st two equation by differentiating them and seeing whether they are constant or not as commented above
A: If we put $t=2x$ and if prime denotes derivative with respect to $t$ then it is easy to see that we have $f''+f=0$ and this has a unique solution $f= f(0)\cos t + f'(0)\sin t$ and note that given conditions imply that $f(0)=1,f'(0)=1/2$ so that $f(x) = \cos 2x +(1/2)\sin 2x$.
