No Problemmo #1

From $|\ln x| = \sin^6 (2\pi x)$ , we note that the RHS is bounded $0 \le \sin^6 (2\pi x) \le 1$ . Therefore, the range of the LHS where the solutions lie is $0 \le |\ln x| \le 1$ $\implies -1 \le \ln x \le 1$ $\implies \dfrac 1e \le x \le e$ . The graph $|\ln x|$ starts at 1 at $x = \dfrac 1e$ , reduces to 0 at $x=1$ and then increases and ends at 1 at $x=e$ .

While $\sin (2\pi x)$ has a period of 1, because it is squared, its negation half cycles turn into positive. Therefore, the period of $\sin^2 (2\pi x)$ as well as $\sin^6 (2\pi x)$ is $\frac 12$ . For $x$ ranges from $\dfrac 1e \approx 0.3678$ to $e \approx 2.7183$ , $\sin^6 (2\pi x)$ completes $\left \lfloor \dfrac {e - \frac 1e}{\frac 12} \right \rfloor = 4$ cycles.

Since $|\ln x|$ cuts every cycle of $\sin^6 (2\pi x)$ at two points, it cuts the RHS at 8 points. We note that at $x=1$ , $|\ln x| = \sin^6 (2\pi x) = 0$ . Therefore, the equation has $\boxed{9}$ solutions.

We shall use just a bit of graph transformation... We transform the graph from y=lnx to y=|lnx| The draw the graph of (sin 2πx)^6. Be attentive the the curves get narrower as we increase the power of sin. 7 solutions are apparent. For more solutions, we must differentiate and check. lnx is 1 at x= 2.71(approx) and sin2πx is 1 at x=2.75. So there is no solution in the region (2.5, 2.75) We have to check when x tends to 1, Slope of y= (sin2πx)^6=12πsin(2πx)^5(cos2πx) Which is equal to 0 at x=1, and slope of lnx is 1 when x=1 this implies y = lnx intersects the other curve at 2 more points near x=1. Also there might be a solution in the region (0.25,0.5) but | lnx | =1 at x = 1/e and y= sin(2πx)^6 is less than 1 at x=1/e. (Sin2πx)^6 is 1 at x = 0.25. so no solution is possible there. So finally we have 9 solutions.