11.6 Find $\sin \left(\frac{\pi}{2}\left(2 x^{2}+1\right)\right)+\cos \left(\pi\left(x^{2}+4 x\right)\right)=0$.
Answer: $-1+\sqrt{\frac{3}{2}}$.

Solution. Writing the first term as $\cos \pi x^{2}$ and transforming the sum of cosines into a product, we obtain two series of roots $4 x=2 n+1$ and $2 x^{2}+4 x=2 m+1$ (where $m, n$ are integers). The positive roots are of the form $x=\frac{n}{2}+\frac{1}{4}$ and $x=-1+\sqrt{m+\frac{3}{2}}$ (where $m, n$ are non-negative integers). For these series, the smallest roots are $\frac{1}{4}$ and $-1+\sqrt{\frac{3}{2}}<\frac{1}{4}$ (the last inequality is verified by squaring both sides of the equivalent inequality $\sqrt{\frac{3}{2}}<\frac{5}{4}$ ).