Problem 3. Determine how many roots of the equation

$$
4 \sin 2 x + 3 \cos 2 x - 2 \sin x - 4 \cos x + 1 = 0
$$

are located on the interval $\left[10^{2014!} \pi ; 10^{2014!+2015} \pi\right]$. In your answer, write the sum of all digits of the found number.
Solution. Let $t=\sin x+2 \cos x$. Then $t^{2}=\sin ^{2} x+2 \sin 2 x+4 \cos ^{2} x=2 \sin 2 x+\frac{3}{2} \cos 2 x+\frac{5}{2}$. The original equation is equivalent to $t^{2}-t-2=0$. From which

$$
\left[\begin{array}{c}
\operatorname{sin} x + 2 \operatorname{cos} x = -1, \\
\operatorname{sin} x + 2 \operatorname{cos} x = 2
\end{array} \Longleftrightarrow \left[\begin{array}{c}
\sin (x+\varphi)=-\frac{1}{\sqrt{5}} \\
\sin (x+\varphi)=\frac{2}{\sqrt{5}}
\end{array}\right.\right.
$$

where $\varphi=\operatorname{arctg} 2$. Therefore, the original equation has exactly four roots on any interval of the form $[2 k \pi ; 2(k+1) \pi], k \in \mathbb{Z}$. Let the number of roots be denoted by $N$. Since there are four different roots in one period (the minimum period of the functions involved in the equation is $2 \pi$), then

$$
\begin{aligned}
& N=4 \cdot \frac{10^{2014!+2015} \pi-10^{2014!} \pi}{2 \pi}=2 \cdot 10^{2014!}\left(10^{2015}-1\right)= \\
&=2 \cdot 10^{2014!} \cdot \underbrace{9 \ldots 9}_{2015}=1 \underbrace{9 \ldots 9}_{2014} 8 \cdot 10^{2014!}=1 \underbrace{9 \ldots 9}_{2014} 8 \underbrace{0 \ldots 0}_{2014!}
\end{aligned}
$$

The sum of all digits of this number is $1+8+9 \cdot 2014=9 \cdot 2015=18135$.

## Answer: 18135