<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] We use $$\sin A+\sin B=2\sin \left( \frac {A+B}{2}\right) \cos \left(\frac {A-B}{2}\right)$$ and $$\cos A+\cos B=2\cos \left( \frac {A+B}{2} \right) \cos \left( \frac {A-B}{2} \right)$$
Hence, $$(\sin x+\sin 3x)-3\sin 2x=2\sin 2x \cos x-3\sin 2x=\sin 2x(2\cos x-3)$$
Moreover, $$(\cos x+\cos 3x)-3\cos 2x=2\cos 2x \cos x-3\cos 2x=\cos 2x(2\cos x-3)$$
Thus, the left hand side of your equation is-
$$\sin 2x(2\cos x-3)-\cos 2x(2\cos x-3)=(\sin2x-\cos 2x)(2\cos x-3)$$
This has been equated to $0$, hence solutions can be found by solving $\cos x=\frac 32$ or $\tan 2x=1$. Note that $\cos x=\frac 32$ has no real solutions.
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