anonymous 5 years ago I have a problem that I am stumped on. Solve for x. Sin(x)=Sin*2(x)

1. anonymous

Ok. 0 squared is zero and 1 squared is 1. So where the sine function is 0 and 1 would satisfy the equation. Where is sine 0 and where is sine 1?

2. anonymous

0pi+2kpi, pi/2+2kp i

3. anonymous

Basically what I'm saying is that the sine function allows you to get 0 and 1 because the sine function varies from -1 to 1. Since you have to have the function itself equal it's square, you have to obtain 0 and 1 to solve your equation.

4. anonymous

What about $\pi$ and (3/2)$\pi$?

5. anonymous

6. anonymous

pi would work though for sinx=0

7. anonymous

Yes, but (3/2)pi would not. That would give you a negative 1, which doesn't satisfy the equation.

8. anonymous

sine of pi is 1. so yes it would work. but 3pi/2 would not since it gives you negative 1 for sin(x), where as sin(3pi/2) squared would be 1 since -1 squared is 1

9. anonymous

Yes, I got that.

10. anonymous

11. anonymous

12. anonymous

so my final answer would be 0pi+2kpi, pi+2kpi, and (pi/2)+2kpi?

13. anonymous

That's right, but the first two can be combined to one. The last one is correct.

14. anonymous

just 0pi+kpi?

15. anonymous

Yes. :) Technically you don't need the 0pi, but it doesn't hurt.

16. anonymous

Thank you very much. You both have been a big help!

17. anonymous

No problem.