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

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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

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

$=> sin(x) = - 1$

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

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

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