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factor the sin^2(x)+sin(x)
there is a sin(x) factor in each term
sin x (sin x + 1) = 0 ?
yes now set both factors equal to 0 sin(x)=0 or sin(x)+1=0
so you have to solve both of those equations sin(x)=0 or sin(x)=-1
sin x = 0 , 2pi sin x = -1 would be x = 3pi/2 ?
what about x=pi or x=0 sin(0)=? sin(pi)=?
i get very confused by the unit circle :/
When you look at the unit circle you are looking for when the y-coordinate is 0 since we are talking about sine cos is the x-coordinate
that is when we are solving sin(x)=0
you already found all the solutions to sin(x)=-1 in the interval [0,2pi]
3pi/2 is the only solution to sin x = -1 ?
sin(x)=0 when x=0,pi,2pi those are the solutions in [0,2pi] There are infinitely many more solutions outside that interval
yes sin(x)=-1 has solution x=3pi/2 only if we are looking at [0,2pi]
As I said though if we aren't restricted to this interval then there are infinitely many solutions.
okay, i understand now.
Your question says to find all the solutions.
so it seems we are restricted to the interval [0,2pi]
what 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
multiples of 2pi right?
sin(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
well to the first equation sin(x)=0 the answers are multiples of pi
you 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
so you do have to include all the solutions to sin(x)=0 we say the solutions are x=npi where n is an integer
okay, i see now