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Ta có: $\sin^3x+\cos^3x+2\sin^2x=1$$\Leftrightarrow \sin^3x+\cos^3x=\cos^2x-\sin^2x$$\Leftrightarrow (\cos x+\sin x)(\sin^2x-\cos x\sin x+\cos^2x)=(\cos x+\sin x)(\cos x-\sin x)$$\Leftrightarrow (\cos x+\sin x)(1+\sin x-\cos x-\cos x\sin x)=0$$\Leftrightarrow (\cos x+\sin x)(1-\cos x)(1+\sin x)=0$$\Leftrightarrow \left[ \begin{array}{l} \tan x=-1\\\sin x=-1\\\cos x=1 \end{array} \right.\Leftrightarrow \left[\begin{array}{l} x=-\frac{\pi}{4}+k\pi\\x=-\frac{\pi}{2}+k2\pi\\x=k2\pi \end{array} \right.,k\in\mathbb{Z}$.

Ta có: $\sin^3x+\cos^3x+2\sin^2x=1$$\Leftrightarrow \sin^3x+\cos^3x=\cos^2x-\sin^2x$$\Leftrightarrow (\cos x+\sin x)(\sin^2x-\cos x\sin x+\cos^2x)=(\cos x+\sin x)(\cos x-\sin x)$$\Leftrightarrow (\cos x+\sin x)(1+\sin x-\cos x-\cos x\sin x)=0$$\Leftrightarrow (\cos x+\sin x)(1-\cos x)(1+\sin x)=0$$\Leftrightarrow \left[ \begin{array}{l} \tan x=-1\\\sin x=-1\\\cos x=1 \end{array} \right.\Leftrightarrow \left[\begin{array}{l} x=-\frac{\pi}{4}+k\pi\\x=-\frac{\pi}{2}+k2\pi\\x=k2\pi \end{array} \right.,k\in\mathbb{Z}$.