Solving equations of the form $c_0 \sin\left(x+p_0\right)+c_1\sin\left(2x+p_1\right)=0$ How do I solve this equation?
$$c_0 \sin\left(x+p_0\right)+c_1\sin\left(2x\right)=0$$
where $c_0$, $p_0$, and $c_1$ are fixed constants? What about the more general form:
$$c_0 \sin\left(x+p_0\right)+c_1\sin\left(2x+p_1\right)=0$$
where $c_0$, $p_0$, $c_1$, and $p_1$ are constants?
 A: WLOG, $c_1=1$.
The equation can be rewritten under the form 
$$s \sin x+c\cos x+\sin x\cos x=0.$$
Then we use the substitution
$$\sin x=\frac{2t}{t^2+1},\cos x=\frac{t^2-1}{t^2+1}$$ 
and obtain the quartic equation
$$t^4+2\frac{s+1}ct^3+2\frac{s-1}ct-1=0.$$
I don't think that there will be a simplification in the expressions of the roots and you must solve this using the general formulas. There will be two or four real roots in $t$, i.e. an even number of roots per period, as was expected.

By the theory of quartic equations, the number of roots is two or four depending on the sign of the discriminant
$$-27(b^4+d^4)-4(bd)^3+6(bd)^2-192bd-256$$ where $b$ and $d$ are the cubic and linear coefficients above.

Letting $u:=\cos x$ and $v:=\sin x$, the given equation is that of an equilateral hyperbola, which passes through the origin,
$$(u+s)(v+c)=cs,$$ and we are intersecting it with the unit circle
$$u^2+v^2=1.$$

A: Original post
If either $c_0=0$ or $c_1=0$ (or both) the problem is trivial.
Hence we assume the opposite and let $a=\frac{c_1}{c_0}$. Furthermore we let $x=y-p_{1}/2$ and $q=p_{0}-p_{1}/2$. Then your generalized equation becomes
$$ \sin(y+q) + a \sin(2y) = 0\tag{1}$$
Since $\sin(y+q) = \sin(y) \cos(q) + \cos(y) \sin(q)$ and $\sin(2y) = 2 \sin(y) \cos(y)$  we get
$$ \sin(y) \cos(q) +  \cos(y) \sin(q) +  2 a \sin(y) \cos(y) = 0\tag{2}$$
Now letting $z = \sin(y)$ we have $\cos(y) = \sqrt{1-z^2}$ and $(2)$ becomes
$$ z \cos(q) +  \sqrt{1-z^2} \sin(q) +  2 a z \sqrt{1-z^2}  = 0\tag{3}$$
This results is an algebraic equation of 8th degree in $z$. But it depends only on $z^2$ so that it is 4th degree in $s=z^2$. Let the four solutions be $s_{k}$, $k=1..4$.
Then the solutions $y$ are given by $y_{k}= \arcsin(\sqrt{s_{k}})$, $k=1..4$
EDIT
Here are some plots of the LHS of $(1)$ for $x \in [0,2\pi)$ for randomly chosen parameters $q\in (0,2\pi)$ and $a\in(-2,2)$

There are 2 roots (figs. 6,8,9), 3 roots (two have merged into a double root, figs. 5 and 7), in all other cases there are 4 roots.
