1. ## sinx = x/100

How many solutions does this equation have: sinx = x/100

2. Originally Posted by Garas
How many solutions does this equation have: sinx = x/100

Clearly we only need to consider $\displaystyle -100 \le x \le 100$. For each period of sin(x), the line y=x/100 will cross the curve y=sin(x) twice (starting at the origin), except that next to the origin it will only cross once, and be careful at the endpoints. Graph it if you are unsure.

In case you didn't follow what I just wrote: On the interval (0, 2pi] there is one intersection, on the interval [2pi, 4pi] there are two intersections, etc. On the interval [-2pi, 0) there is one intersection, on the interval [-4pi, -2pi] there are two intersections, etc.

And don't forget the intersection at the origin.

Also note that due to symmetry the number of intersections in (0,100] will be the same as the number of intersections in [-100,0).

3. I know that but i'm not sure how many times does those two graphs cross each other.

4. Originally Posted by Garas
I know that but i'm not sure how many times does those two graphs cross each other.
Re-read what I wrote. I made a few edits. If you understand what I wrote, then you know how to answer the question.

5. I need correct number in order to check my solution

6. Originally Posted by Garas
I need correct number in order to check my solution
Show some work, or ask a specific question. I am not a homework machine.

7. My specific question is: how many solutions does equation have?

8. Originally Posted by Garas
My specific question is: how many solutions does equation have?
You wrote "I need correct number in order to check my solution."

When I said to ask a specific question, I meant a specific question on how to do a certain step, or a question about my explanation if you didn't understand it. This was implied.

9. 1+4*floor(100/2pi)

10. Originally Posted by Garas
1+4*floor(100/2pi)
Thank you! Now we're getting somewhere.

Your answer is slightly off. floor(100/2pi) = 15, but in the first period we have 2 intersections instead of 4, and in the partial period near the endpoints we have 4 intersections. So it's 1+(2*1)+(4*15)=63. Another way of doing it is 1+2*(1+2*15)=63. Here's an image I made with Mathematica. Your formula gives the incorrect answer 61.

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# sinX= x/10 how many solutions

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