Question 41, If real numbers $x, y$ satisfy $\left\{\begin{array}{c}x^{3}+\cos x+x-2=0 \\ 8 y^{3}-2 \cos ^{2} y+2 y+3=0\end{array}\right.$, then $\cos (x+2 y)=$
Question 41, Answer: Transforming the equation, we get $\left\{\begin{array}{c}x^{3}+\cos x+x-2=0 \\ f(t) \\ (-2 y)^{3}+\cos (-2 y)+(-2 y)-2=0^{\circ}\end{array}\right.$ Let $f(t)=t^{3}+\cos t+t-2$, then $f(x)=f(-2 y)$. Since $f(t)=3 t^{2}-\sin t+1>0$, this indicates that $f(t)$ is monotonically increasing, hence $f(t)$ has only one zero point, so $x=-2 y$, which means $x+2 y=0$. Therefore, $\cos (x+2 y)=1$