$f(x)=a\sin x + b \cos x\equiv 0$ 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?

Your assumption that $a$ is $0$ is wrong. For $a\ne 0$ the result is true. But then we arrive at a contradiction, implying that $a=0\implies b=0$
