Question

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

Answer 1

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?

Answer 2

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

Answer 3

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.

Answer 4

What about \[\pi\] and (3/2)\[\pi\]?

Answer 5

Haha nevermind, misread the question. My bad.

Answer 6

pi would work though for sinx=0

Answer 7

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

Answer 8

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

Answer 9

Yes, I got that.

Answer 10

ok. So your answers were correct I believe.

Answer 11

His answers are close, they need to be adjusted.

Answer 12

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

Answer 13

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

Answer 14

just 0pi+kpi?

Answer 15

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

Answer 16

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

Answer 17

No problem.