53. Solve the equation $\cos 2x = \lg x$.
Solution. Since $D(\cos )=R$ and $D(\mathrm{lg})=R_{+}$, the roots of the equation, if they exist, must be positive.

The roots of the equation cannot be greater than 10, because for $x>10$ $\lg x>1$, while $\cos 2 x$ cannot be greater than one. Therefore, the roots of the given equation, if they exist, are between 0 and 10.

If $0 \leqslant x \leqslant \frac{\pi}{4}$, then $\lg x\lg \frac{\pi}{4}$, and $\cos \left(2 \cdot \frac{\pi}{2}\right)0$.

![](https://cdn.mathpix.com/cropped/2024_05_21_b1a79769c6c63d19f37fg-33.jpg?height=392&width=1328&top_left_y=238&top_left_x=364)

Fig. 26

In the interval $\frac{3 \pi}{4} \leqslant x \leqslant \pi \cos \left(2 \cdot \frac{3 \pi}{4}\right)\lg \pi$. Since the functions $\cos 2 x$ and $\lg x$ are monotonic and continuous, there is some value $x_{2}$ in this interval such that $\cos 2 x_{2}=\lg x_{2}$.

In the interval $\pi \leqslant x \leqslant \frac{5 \pi}{2} \cos (2 \cdot \pi)>\lg \pi$ and $\cos \left(2 \cdot \frac{5 \pi}{4}\right)0$.

By similar reasoning, we obtain two roots: $x_{4}$ and $x_{5}$, belonging to the interval $\left[\frac{7 \pi}{4} ; \frac{9 \pi}{4}\right]$, and two roots: $x_{6}$ and $x_{i}$, belonging to the interval $\left[\frac{11 \pi}{4} ; \frac{13 \pi}{4}\right]$.

Since $\frac{13 \pi}{4}>10$, there are no more roots in the interval $\left[\frac{13 \pi}{4} ;+\infty\right)$.

Therefore, our equation has seven solutions.

By plotting the graphs of the functions $\cos 2 x$ and $\lg x$ on graph paper (Fig. 26), we find them: $x_{1} \approx 0.8, x_{2} \approx 2.4, x_{3} \approx 3.6, x_{4} \approx 5.7, x_{5} \approx 6.7$, $x_{6} \approx 8.9, x_{7} \approx 9.8$