4B. If $\cos x+\cos ^{2} x+\cos ^{3} x=1$, calculate the value of the expression

$$
\sin ^{6} x-4 \sin ^{4} x+8 \sin ^{2} x
$$
Solution. From $\cos x+\cos ^{2} x+\cos ^{3} x=1$ we sequentially obtain:

$$
\begin{aligned}
& \left(\cos x+\cos ^{3} x\right)^{2}=\left(1-\cos ^{2} x\right)^{2} \\
& \cos ^{2} x+2 \cos ^{4} x+\cos ^{6} x=\sin ^{4} x \\
& 1-\sin ^{2} x+2\left(1-\sin ^{2} x\right)^{2}+\left(1-\sin ^{2} x\right)^{3}=\sin ^{4} x \\
& 1-\sin ^{2} x+2-4 \sin ^{2} x+2 \sin ^{4} x+1-3 \sin ^{2} x+3 \sin ^{4} x-\sin ^{6} x=\sin ^{4} x \\
& 4-8 \sin ^{2} x+4 \sin ^{4} x-\sin ^{6} x=0 \\
& \sin ^{6} x-4 \sin ^{4} x+8 \sin ^{2} x=4
\end{aligned}
$$

## 4th Year