$\sin^3x+2\sin x\cos^2x+3\cos^3x=0$ (*)
- Với $\cos x=0$, thay vào (*) ta có:
$\sin^3x=0$ (loại)
- Với $\cos x\ne 0\to x\ne \dfrac{\pi}{2}+k\pi$
Chia hai vế (*) cho $\cos^3x$ ta có:
$\tan^3x+2\tan x+3=0$
$\to (\tan x+1)(\tan^2x-\tan x+3)=0$
$\to \left[\begin{matrix} \tan x=-1\\ \tan^2x-\tan x+3=0(VN)\end{matrix}\right.$
$\to x=\dfrac{-\pi}{4}+k\pi$ (TM)