$f(x)=a\sin x + b \cos x\equiv 0$ Let $$f(x)=a\sin x + b \cos x$$
where a,b are some constants.
Let there are exists $x_1, x_2$ such that $$f(x_1)=f(x_2)=0$$ $x_1-x_2\not =\pi k, k \in \mathbb Z $ .
Prove that $f(x)=0$ for all $x\in \mathbb R$. And is it $a=b=0$ in this case or not?
###My work
I see that $f(x_1)=f(x_2)=0$.
$$a\sin x_1 + b \cos x_1=0 \Rightarrow \tan x_1=-\frac ba $$
$$a\sin x_2 + b \cos x_2=0 \Rightarrow \tan x_2=-\frac ba $$
Hence, $\tan x_1-\tan x_2=0$. Then $\sin(x_1- x_2)=0 \Rightarrow x_1-x_2=\pi k$.
But $x_1-x_2\not =\pi k$.
Where my mistake? How I need continue to solve my problem?

Keep $\tan$ out of the game.
When $(a,b)\ne(0,0)$ then the point $(a,b)\in{\mathbb R}^2$ has distance $A:\sqrt{a^2+b^2}>0$ from the origin, and there is a unique $\phi\in[0,2\pi[\>$ such that
$$a=A\cos\phi,\quad b=A\sin\phi\ .$$
It follows that
$$f(x)=A(\cos\phi\sin x+\sin\phi\cos x)=A\sin(x+\phi)\ ,$$
so that $f$ has two zeros differing $\pi$ modulo $2\pi$. Since such pairs of zeros have been excluded it follows that in fact $(a,b)=(0,0)$, so that $f(x)\equiv0$.
