How do you solve #sin^2x+sinx=0# and find all solutions in the interval #[0,2pi)#?

Question

51211 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-23T10:28:57

#x = 0, (3 pi) / (2), pi#

We have: #sin^(2)(x) + sin(x) = 0#; #[0, 2 pi)#

#=> sin(x) (sin(x) + 1) = 0#

#=> sin(x) = 0#

#=> x = 0, (pi - 0), (pi + 0), (2 pi - 0)#

or

#=> sin(x) + 1 = 0#

#=> sin(x) = - 1#

#=> x = pi + (pi) / (2)#

#=> x = 0, (3 pi) / (2), pi#