How many times does the graph of y=0.1x intersect the graph of y=sin(2x)

Check over domain -10<=x<=10 since |sin(2x)| cannot be greater than 1.

Also note that sin(2x) is periodic with a period of pi.

First start at x=0. Note that 0.1(0) = sin(2*0) is an intersection. Also note that sin(2x) will range from -1 to 1 every pi units we travel along the x-axis. Also note that the rate of increase of sin(2x) at x=0 along the positive x direction is greater than that of 0.1x so there will only be one additional intersection along the domain 0<x<=pi rather than 2. (If this last part sounds confusing I can explain in greater detail but you'll have to wait awhile.) Finally note that from the domains pi<x<=2pi and 2pi<x<=3pi we have 2 intersections each. So from 0<=x<=3pi we have a total of 6 intersections.

All that we need to check now is if there are any more intersections on the domain 3pi<x<=10. Sin(2x) will reach its maximum at x=10 on this domain since the function increases on this domain so we have:

sin(2*10) = sin(20) = .913 (approximate)

0.1(3pi) = .942 (approximate)

Since 0.1x is larger than sin(2x) for the entire domain 3pi<x<=10 we can conclude that there is no additional intersections.

Now all that is left to do is note that sin(2x) and 0.1x are both symmetric with respect to the origin so the number of intersections on the domain -10<=x<0 will be the same as the number of intersections on the domain 0<x<=10, which is 5. (I left out x=0 since we counted that one already).

Therefore assuming I haven't messed up somewhere (and assuming x was measured in radians and not degrees) the number of intersections is 6+5=11.

Hope it helps :)

step-by-step explanation:

this is not possible to solve with the info given.   you need at least an one angle somewhere.