Solve $2\sin^3x + \sin3x +3\sin^2x \cos x + \cos^3x=0$ $2\sin^3x + \sin3x +3\sin^2x\cos x + \cos^3x=0$
My try:
$$2\sin^3x +3\sin x - 4\sin^3x +\cos x(3\sin^2x+\cos^2x)=0 $$
$$ \cos x(2\sin^2x+1) - 2\sin^3x+3\sin x=0.$$
And then i have no idea.
 A: Since
$$
\sin 3x=3\cos^2x\sin x-\sin^3x
$$
you actually have
$$
\sin^3x+3\sin^2x\cos x+3\sin x\cos^2x+\cos^3 x=0,
$$
or, using the binomial theorem,
$$
(\sin x+\cos x)^3=0.
$$
I'm sure you can take it from here.
A: Notice, $$2\sin^3x + \sin3x +3\sin^2x\cos x + \cos^3x=0$$
$$2\sin^3x +3\sin x - 4\sin^3x +3\cos x(1-\cos^2x)+cos^3x=0 $$
$$3\sin x - 2\sin^3x +3\cos x-3\cos^3x+cos^3x=0 $$
$$ 3(\cos x+\sin x)-2(\cos^3x+\sin^3x)=0$$ $$ 3(\cos x+\sin x)-2(\cos x+\sin x)(\cos^2x+\sin^2x-\cos x\sin x)=0$$ $$(\cos x+\sin x)(3-2(1-\cos x\sin x))=0$$$$\implies \cos x+\sin x=0\implies \frac{1}{\sqrt{2}}\cos x+\frac{1}{\sqrt{2}}\sin x=0$$ $$\implies \cos\left(x-\frac{\pi}{4}\right)=0 \implies x-\frac{\pi}{4}=(2n+1)\frac{\pi}{2}$$ $$\implies x=(2n+1)\frac{\pi}{2}+\frac{\pi}{4}$$$$\implies \color{blue}{x=\frac{(4n+3)\pi}{4}}$$ or $$\implies 3-2(1-\cos x\sin x)=0\implies 2\sin x\cos x=-1\implies 2x=2n\pi-\frac{\pi}{2}$$ $$\implies x=n\pi-\frac{\pi}{4}=\color{blue}{\frac{(4n-1)\pi}{4}}$$ Where $\color{blue}{n}$ is any integer
