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anonymous
 one year ago
Find all solutions to the equation.
sin^2 x + sin x = 0
anonymous
 one year ago
Find all solutions to the equation. sin^2 x + sin x = 0

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freckles
 one year ago
Best ResponseYou've already chosen the best response.1factor the sin^2(x)+sin(x)

freckles
 one year ago
Best ResponseYou've already chosen the best response.1there is a sin(x) factor in each term

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sin x (sin x + 1) = 0 ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1yes now set both factors equal to 0 sin(x)=0 or sin(x)+1=0

freckles
 one year ago
Best ResponseYou've already chosen the best response.1so you have to solve both of those equations sin(x)=0 or sin(x)=1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sin x = 0 , 2pi sin x = 1 would be x = 3pi/2 ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1what about x=pi or x=0 sin(0)=? sin(pi)=?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i get very confused by the unit circle :/

freckles
 one year ago
Best ResponseYou've already chosen the best response.1When you look at the unit circle you are looking for when the ycoordinate is 0 since we are talking about sine cos is the xcoordinate

freckles
 one year ago
Best ResponseYou've already chosen the best response.1that is when we are solving sin(x)=0

freckles
 one year ago
Best ResponseYou've already chosen the best response.1you already found all the solutions to sin(x)=1 in the interval [0,2pi]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.03pi/2 is the only solution to sin x = 1 ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1sin(x)=0 when x=0,pi,2pi those are the solutions in [0,2pi] There are infinitely many more solutions outside that interval

freckles
 one year ago
Best ResponseYou've already chosen the best response.1yes sin(x)=1 has solution x=3pi/2 only if we are looking at [0,2pi]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1As I said though if we aren't restricted to this interval then there are infinitely many solutions.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay, i understand now.

freckles
 one year ago
Best ResponseYou've already chosen the best response.1Your question says to find all the solutions.

freckles
 one year ago
Best ResponseYou've already chosen the best response.1so it seems we are restricted to the interval [0,2pi]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1what did you notice about the solutions to sin(x)=0 like we had solutions 0pi,1pi,2pi,...continuing this pattern you should be able to say in general the solutions to sin(x)=0 is npi where n is an integer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0multiples of 2pi right?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1sin(x)=1 you said had solution x=3pi/2 to include all the solutions you could recall the period of sin is 2pi and say the solutions to sin(x)=1 are x=3pi/2+2pi*n where again n is an integer

freckles
 one year ago
Best ResponseYou've already chosen the best response.1well to the first equation sin(x)=0 the answers are multiples of pi

freckles
 one year ago
Best ResponseYou've already chosen the best response.1you say the solution is sin(x)=0 is x=2npi you are excluding odd n..because the 2 in front of the n makes it is even for any n for example x=2npi doesn't include the solution x=pi

freckles
 one year ago
Best ResponseYou've already chosen the best response.1so you do have to include all the solutions to sin(x)=0 we say the solutions are x=npi where n is an integer
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