$8.469 \sin ^{4} x+2 \cos ^{3} x+2 \sin ^{2} x-\cos x+1=0$.

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$8.469 \sin ^{4} x+2 \cos ^{3} x+2 \sin ^{2} x-\cos x+1=0$.
Solution. $\sin ^{4} x+2 \sin ^{2} x+1=\cos x \cdot\left(1-2 \cos ^{2} x\right) \Leftrightarrow\left(\sin ^{2} x+1\right)^{2}=-\cos x \cdot \cos 2 x$.

For any $x\left(\sin ^{2} x+1\right)^{2} \geq 1,-\cos x \cdot \cos 2 x \leq 1$. Therefore $\left\{\begin{array}{l}\sin ^{2} x+1=1, \\ -\cos x \cdot \cos 2 x=1\end{array} \Leftrightarrow\left\{\begin{array}{l}\sin x=0, \\ -\cos x \cdot\left(1-2 \sin ^{2} x\right)=1\end{array} \Leftrightarrow\left\{\begin{array}{l}\sin x=0, \\ \cos x=-1\end{array} \Leftrightarrow x=\pi+2 \pi k\right.\right.\right.$.

Answer: $\quad x=\pi(2 k+1)$.

$8.470|\sin x|+|\cos x|=1.4$.

Solution. Square both sides of the equation:

$|\sin 2 x|=\frac{24}{25} \Leftrightarrow \sin 2 x= \pm \frac{24}{25} \Leftrightarrow x= \pm \frac{1}{2} \arcsin \frac{24}{25}+\frac{\pi k}{2}$.

Answer: $x= \pm \frac{1}{2} \arcsin \frac{24}{25}+\frac{\pi k}{2}$.