<s>[INST] How to solve $ \sin(x) - \cos(x) -3\sin(2x) +3\cos(2x) +\sin(3x) -\cos(3x) =0$ Hello friends please help me to solve this equation.
$ \sin(x) - \cos(x) -3\sin(2x) +3\cos(2x) +\sin(3x) -\cos(3x) =0.$


Here is what I have done so far

$ \Rightarrow \sqrt2(\frac{1}{\sqrt2}\sin x - \frac{1}{\sqrt2}\cos x ) - 3\sqrt2(\frac{1}{\sqrt2}\sin( 2x) - \frac{1}{\sqrt2}\cos (2x) )+ \sqrt2(\frac{1}{\sqrt2}\sin x - \frac{1}{\sqrt2}\cos x )=0$

$\Rightarrow  2(\frac{1}{\sqrt2}\sin x - \frac{1}{\sqrt2}\cos x ) - 3(\frac{1}{\sqrt2}\sin(2x) - \frac{1}{\sqrt2}\cos (2x) )=0$


$\Rightarrow 3\cos(\frac{\pi}{4} + 2x) -2\cos(\frac{\pi}{4}+x) =0$


Please help me to continue this method and I'd there is an easier way please tell me it too.
 [/INST] You changed a couple of $3x$ terms into $x$, so your attempt is faulty.
Consider instead that
$$
\sin 3x+\sin x=2\sin2x\cos x,\qquad
\cos 3x+\cos x=2\cos2x\cos x
$$
so your equation becomes
$$
3(\cos2x-\sin2x)-2\cos x(\cos2x-\sin2x)=0
$$
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