2. Solve the equation $8 \sin ^{4}(\pi x)-\sin ^{2} x=\cos ^{2} x-\cos (4 \pi x)$. In your answer, specify the sum of the roots that belong to the interval $[-1 ; 2]$.

points)
Solution. Considering the basic trigonometric identity, we get

$8 \sin ^{4}(\pi x)-1+\cos (4 \pi x)=0 \quad \Rightarrow \quad 8 \sin ^{4}(\pi x)-2 \sin ^{2}(2 \pi x)=0 \quad \Rightarrow$

$\left(2 \sin ^{2}(\pi x)-2 \sin (\pi x) \cos (\pi x)\right)\left(2 \sin ^{2}(\pi x)+2 \sin (\pi x) \cos (\pi x)\right)=0 \Rightarrow$ $\sin ^{2}(\pi x)\left(\sin ^{2}(\pi x)-\cos ^{2}(\pi x)\right)=0 \Rightarrow \sin ^{2}(\pi x)(\cos (2 \pi x))=0$.

Therefore, $\left[\begin{array}{l}\sin (\pi x)=0, \\ \cos (2 \pi x)=0,\end{array} \Rightarrow\left[\begin{array}{l}\pi x=\pi k, k \in Z, \\ 2 \pi x=\frac{\pi}{2}+\pi n, n \in Z,\end{array} \Rightarrow\left[\begin{array}{l}x=k, k \in Z, \\ x=\frac{1}{4}+\frac{n}{2}, n \in Z,\end{array}\right.\right.\right.$ then the roots of the equation belonging to the interval $[-1 ; 2]$, will be $-1,1,2,-0.75,-0.25,0.25,0.75,1.25,1.75$. Their sum is 5.

## Answer: 5.