Solving $2\cos(x) = \sin(x)$ How would you solve equations of the form $ a \sin (x+b) = \sin (x)$?
Eg. $ 2 \cos(x) = \sin(x) $
I realy have no idea how I would solve this kind of equations.
 A: For your initial question, you can use the angle-sum rule for $\sin(\alpha + \beta):$ $$\sin(\alpha \pm \beta) = \sin \alpha \cos \beta \pm \cos \alpha \sin \beta$$
So $$a \sin (x+b) = \sin (x) \iff a (\sin x \cos b + \cos x \sin b) = \sin x$$
We can divide through by $\cos x$, provided $\cos x \neq 0$, noting that $a, \cos b, \sin b$ are all constants.

For your example:
Note that if $\cos x \neq 0$, then $2 \cos x = \sin x \iff 2 = \dfrac {\sin x}{\cos x} = \tan x \iff \tan^{-1}(2) = x$.
And as @Alex notes,  $2\cos x = \sin x \implies \cos x \neq 0$.
A: I would do a tan half angle substitution with $t = \tan \frac{x}{2}$ which leads to
 $$ \cos(x) = \cos\left( 2 \tan^{-1} t \right) = \frac{1-t^2}{1+t^2} \\
\sin(x) =\sin \left( 2 \tan^{-1} t \right) =\frac{2 t}{1+t^2} $$
which transforms your problem to
$$ \frac{1-t^2}{1+t^2} a \sin(b) + \frac{2 t}{1+t^2} \left(a \cos(b)-1\right) =0 \rightarrow$$
$$ 2 t \left(a \cos b -1\right) + a (1-t^2)\sin b=0  $$
The above has two solutions
$$ \boxed{ t = \dfrac{a\cos b-1 \pm \sqrt{c}}{a \sin b} }$$ where $c=1+a (a-2\cos b) $, and $ \boxed{x= 2 \tan^{-1} t }$.
So with $a=2$ and $b=0$ the result is $x =\pm \pi$
Notes:
This method allows the transformation of any trigonometric expression into a polynomial expression. it is often used in Robotics.
A: $$
a\sin(x+b) = a\sin x\cos b + a\sin b\cos x = \sin x \\\\
\implies (a\cos b - 1)\sin x = a\sin b\cos x \\\\
\implies \tan x = \frac{a \sin b}{a\cos b - 1}\\\\
\implies x = \arctan\left(\frac{a \sin b}{a\cos b - 1}\right)$$
A: $$2\cos(x)- \sin(x)=0$$
$$-2\cos(x)+\sin(x)=0$$
$$\sqrt{5}*\sin(x-\arctan{2})=0$$
You can solve it also in this way.
