Let $f(x)=1+2\cos x +3 \sin x$. If real number $a, b, c$ are such that $af(x)+bf(x+c)=1$ holds for any $x \in \mathbb R$, then find $a, b, c$ 

Let $f(x)=1+2\cos x +3\sin x$. If real number $a, b, c$ are such that $af(x)+bf(x+c)=1$ holds for any $x \in \mathbb R$, then find $a, b, c$


Solution given as in book:
since $f(x)+f(x+\pi)=2 \implies \dfrac{1}{2}f(x)+\dfrac{1}{2}f(x+\pi)=1$
and after comparing with original equation we get $a=\dfrac{1}{2}, b=\dfrac{1}{2}, c=\pi$
My doubt: Isn't there other way to approach this problem as it is not obvious to do  $f(x)+f(x+\pi)=2$ because this kind of step does not come to mind immediately.
 A: Using $\cos(\theta)=\sin\left(\theta+\frac{\pi}{2}\right)$, as well as several formulas for linear combinations of $\sin$ and $\cos$ (as suggested in Hypernova's comment), gives
$$\begin{equation}\begin{aligned}
1 &= af(x)+bf(x+c) \\
& = a(1+3\sin(x)+2\cos(x)) + b(1+3\sin(x+c)+2\cos(x+c)) \\
& = a + b + a\sqrt{13}\cos(x+\alpha) + b\sqrt{13}\cos(x+c+\alpha) \\
& = (a + b) + \sqrt{13}\left(a\sin\left(x+\alpha+\frac{\pi}{2}\right)+b\sin\left(x+\alpha+\frac{\pi}{2}+c\right)\right) \\
& = (a + b) + \sqrt{13}d\sin\left(x+\alpha+\frac{\pi}{2}+\beta\right)
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
where $\alpha=\arctan\left(-\frac{3}{2}\right)$, $d^2=a^2+b^2+2ab\cos(c)$ and $\tan(\beta)=\frac{b\sin(c)}{a+b\cos(c)}$. Because \eqref{eq1A} must be true for all $x$, but with $\sin\left(x+\alpha+\frac{\pi}{2}+\beta\right)$ varying between $-1$ and $1$, we therefore require
$$a+b=1, \; d = 0 \tag{2}\label{eq2A}$$
Next, we have
$$0 \le (a-b)^2-d^2 = -2ab(1+\cos(c)) \tag{3}\label{eq3A}$$
$$1 = (a+b)^2-d^2 = 2ab(1-\cos(c)) \tag{4}\label{eq4A}$$
From \eqref{eq4A}, since $1 - \cos(c) \ge 0$, then $ab \gt 0$. Thus, using \eqref{eq3A}, $1 + \cos(c) \le 0$ so, with $1 + \cos(c) \ge 0$, we have
$$1 + \cos(c) = 0 \; \; \to \; \; \cos(c) = -1 \; \; \to \; \; c = (2k+1)\pi \tag{5}\label{eq5A}$$
for any integer $k$. This then means, from \eqref{eq3A}, that
$$0 = (a-b)^2 - d^2 \; \; \to \; \; a = b \tag{6}\label{eq6A}$$
Finally, we get
$$a + b = 1 \; \; \to \; \; a = b = \frac{1}{2} \tag{7}\label{eq7A}$$
A: Note that $3\sin x+2\cos x=\sqrt{13}\sin (x+\alpha)$ where $\tan \alpha=\frac{2}{3}$.
Now we get $$a+b+\sqrt{13}(a\sin(x+\alpha)+b\sin(x+c+\alpha))=1$$
for all $x\in \mathbb R$.
Now if we set $x=-\alpha$ and $x=\pi-\alpha$, then $a+b=1$ and $\sin c=0$ so $c=k\pi$ for some integer $k$.
Plugging $x=\frac{\pi}{2}-\alpha$ gives $a+b\cos k\pi=0$ so $k$ is odd and $a-b=0$. Therefore $a=b=\frac{1}{2}$.
