ĐK: $x \ne k\pi$
Phương trình đã cho tương đương với:
$2\sin^2x+\cos x=4\sin^2x.\cos x+\sin x$
$\Leftrightarrow \sin x(2\sin x-1)-\cos x(4\sin^2x-1)=0$
$\Leftrightarrow (2\sin x-1)[\sin x-\cos x(2\sin x+1)]=0$
$\Leftrightarrow \left[\begin{array}{l}2\sin x-1=0&(1)\\\sin x-\cos x(2\sin x+1)=0&(2)\end{array}\right.$
$(1) \Leftrightarrow \sin x=\dfrac{1}{2} \Leftrightarrow \left[\begin{array}{l}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{array}\right.,k\in\mathbb{Z}$
$(2) \Leftrightarrow \sin x-\cos x-2\sin x\cos x=0$
Đặt $t=\sin x-\cos x=\sqrt2\sin(x-\dfrac{\pi}{4}),|t|\le\sqrt2$
Khi đó $2\sin x\cos x=1-t^2$
PT $(2)$ trở thành: $t^2+t-1=0$
$\Leftrightarrow \left[\begin{array}{l}t=\dfrac{-1+\sqrt5}{2}\\t=\dfrac{-1-\sqrt5}{2}\end{array}\right.$
Với $t=\dfrac{-1-\sqrt5}{2}$, loại vì $|t|\le\sqrt2$
Với $t=\dfrac{-1+\sqrt5}{2}$, ta có:
$\sin(x-\dfrac{\pi}{4})=\dfrac{-1+\sqrt5}{2\sqrt2}$
$\Leftrightarrow \left[\begin{array}{l}x=\dfrac{\pi}{4}+\arcsin(\dfrac{-1+\sqrt5}{2\sqrt2})+k2\pi\\x=\dfrac{5\pi}{4}+\arcsin(\dfrac{-1+\sqrt5}{2\sqrt2})+k2\pi\end{array}\right.,k\in\mathbb{Z}$