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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?
0pi+2kpi, pi/2+2kp i
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.
What about \[\pi\] and (3/2)\[\pi\]?
Haha nevermind, misread the question. My bad.
pi would work though for sinx=0
Yes, but (3/2)pi would not. That would give you a negative 1, which doesn't satisfy the equation.
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
Yes, I got that.
ok. So your answers were correct I believe.
His answers are close, they need to be adjusted.
so my final answer would be 0pi+2kpi, pi+2kpi, and (pi/2)+2kpi?
That's right, but the first two can be combined to one. The last one is correct.
Yes. :) Technically you don't need the 0pi, but it doesn't hurt.
Thank you very much. You both have been a big help!