Show linear independence of $\{1, \cos x, \sin x\}$ Linear independence of $f:\mathbb{R}\to\mathbb{R}$ given as
$$a+\cos(x)b+\sin(x)c=0$$
First I choose $x=-\frac{\pi}{2}$. I get
$$a+0b-1c=0\iff a=c$$
Then I choose $x=\pi$. I get
$$a-b+0c=0\iff a=b$$
We have that events $(\cos(x)=0\cap \sin(x)=0)=\emptyset$ as they are disjoint events, so we cannot conclude yet that $a=0$ must be true. If we use that $a=b=c$, we have:
$$a+\cos(x)b+\sin(x)c=a+\cos(x)a+\sin(x)a=0\iff 1+\cos(x)+\sin(x)=0\iff x=\pi\vee x=-\frac{\pi}{2}$$
For $x\neq\pi$ and $x\neq-\frac{\pi}{2}$ then $a=b=c=0$ for the equation to be true.
Is this a proper way of showing that they are linear independent?

Just to give a different approach, suppose $a+b\cos x+c\sin x=0$ for all $x$.  Without loss of generality, we can assume $b\ge0$.  If $b^2+c^2\not=0$, then 
$$x=\arcsin\left({c\over\sqrt{b^2+c^2}}\right)\implies a+\sqrt{b^2+c^2}=0$$
while
$$x=\arcsin\left({c\over\sqrt{b^2+c^2}}\right)+\pi\implies a-\sqrt{b^2+c^2}=0$$
which together imply $b^2+c^2=0$, a contradiction. So we must have $b=c=0$, which implies $a=0$ as well.
