To show the zeros of functions are distinct Show that the zeros of the functions $a \sin(x) + b \cos(x)$ and $c \sin(x)  + d \cos(x)$ are distinct and occur alternately if $ad-bc \ne 0$.
I guess I need to use Sturm separation theorem. I need to show that both these functions are linearly independent. This gives me the hint to use wronskian. Is this much enough to show this?
 A: Here a elementary proof by contrapositive: Suppose that for some $x_0$ we have
$$a\sin x_0+b\cos x_0=0\hspace{.4cm}\text{ and }\hspace{.4cm}c\sin x_0+d\cos x_0=0.$$
In terms of the rotation matrix $R_{x_0}=\left(\begin{array}{cc}\cos x_0&-\sin x_0\\ \sin x_0&\cos x_0\end{array}\right)$ this means that
$$
R_{x_0}\binom{a}{b}=\binom{\alpha}{0}\hspace{.4cm}\text{ and }\hspace{.4cm}R_{x_0}\binom{c}{d}=\binom{\beta}{0}
$$
for some $\alpha, \beta$. In particular this says that the images of $\binom{a}{b}$ and $\binom{c}{d}$ under a rotation are collinear, so $\binom{a}{b}$ and $\binom{c}{d}$ must be as well, and then we must have
$$ad-cb=\det\left(\begin{array}{cc}a&c\\ b&d\end{array}\right)=0.$$
Edit: Notice that this argument also works backwards, so one can reword it in order to obtain a direct proof. It just felt more natural to me to express it this way.
A: Well, you don't need anything difficult: just suppose
$$
A = \sqrt{a^2 + b^2}, \qquad  C = \sqrt{c^2 + d^2}:
$$
obviously we have 
$$
 -1 \leq \frac aA \leq 1 , \qquad  -1 \leq \frac bA \leq 1 , \qquad \Big(\frac aA \Big)^2 + \Big(\frac bA \Big)^2 = 1;
$$
and
$$
 -1 \leq \frac cC \leq 1 , \qquad  -1 \leq \frac dC \leq 1 , \qquad \Big(\frac cC \Big)^2 + \Big(\frac dC \Big)^2 = 1.
$$
So you can find two angles $\ \alpha , \ \gamma \ $ with $\ \frac{- \pi}2 \leq \alpha , \ \gamma \leq \frac{ \pi}2 \ $ such that
$$
 \cos(\alpha) = \frac aA , \qquad  \sin(\alpha) = \frac bA , \qquad \cos(\gamma) = \frac cC , \qquad  \sin(\gamma) =\frac dC .
$$
So our functions can be written as
$$
f_1(x) = A\sin (x + \alpha), \qquad  f_2(x) = C\sin (x + \gamma),
$$
and their zeroes can be written explicitely as
$$
x_k = k\pi - \alpha, \qquad  x'_k = k\pi - \gamma.
$$
They are the same only if $\ \alpha = \gamma,\ $ or $\ \alpha = - \gamma = \frac {\pi}2,\ $ or $\ \alpha = - \gamma = -\frac {\pi}2\ $, in which cases your equality is trivial. 
A: We may assume $\sqrt{a^2+b^2}=\sqrt{c^2+d^2}=1$, hence
$$a=\cos\alpha,\quad b=\sin\alpha,\quad c=\cos\beta,\quad d=\sin\beta$$
for certain angles $\alpha$, $\beta$. Then the functions  $$f(x)=a\sin x+b\cos x=\sin(x+\alpha),\qquad g(x):=c\sin x+d\cos x=\sin(x+\beta)$$
have different, and alternating, zeros iff $\beta-\alpha\ne0$ modulo $\pi$, i.e. iff $\sin(\beta-\alpha)=ad-bc\ne0$.
