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

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Answer 1 · 2016-09-23T10:28:57

$x = 0, \frac{3 π}{2}, π$ $x = 0, (3 pi) / (2), pi$

We have: ${sin}^{2} (x) + sin (x) = 0$ $sin^(2)(x) + sin(x) = 0$ ; $[0, 2 π)$ $[0, 2 pi)$

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

$\Rightarrow sin (x) = 0$ $=> sin(x) = 0$

$\Rightarrow x = 0, (π - 0), (π + 0), (2 π - 0)$ $=> x = 0, (pi - 0), (pi + 0), (2 pi - 0)$

or

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

$\Rightarrow sin (x) = - 1$ $=> sin(x) = - 1$

$\Rightarrow x = π + \frac{π}{2}$ $=> x = pi + (pi) / (2)$

$\Rightarrow x = 0, \frac{3 π}{2}, π$ $=> x = 0, (3 pi) / (2), pi$

How do you solve sin^2x+sinx=0sin2x+sinx=0sin^2x+sinx=0 and find all solutions in the interval [0,2pi)[0,2π)[0,2pi)?