Giải thích các bước giải:
$3\sin2x-2\cos3x-10\sin x-3\cos x+6=0$
$\to 3\sin2x-2(4\cos^3x-3\cos x)-10\sin x-3\cos x+6=0$
$\to 6\sin x\cos x-8\cos^3x+3\cos x-10\sin x+6=0$
Đặt $\sin x=a,\cos x=b$
$\to\begin{cases}a^2+b^2=1\\ 6ab-8b^3+3b-10a+6=0\end{cases}$
$\to\begin{cases}a^2+b^2=1\\ a=\dfrac{8b^3-3b-6}{2\left(-5+3b\right)}\end{cases}$
$\to (\dfrac{8b^3-3b-6}{2\left(-5+3b\right)})^2+b^2=1$
$\to \left(8b^3-3b-6\right)^2+4b^2\left(-5+3b\right)^2=4\left(-5+3b\right)^2$
$\to 64b^6-12b^4-216b^3+73b^2+156b-64=0$
$\to b\approx \:0.43603\dots ,\:b\approx \:-0.83819\dots $
$\to x=\arccos b$
