How do you solve the trig equation: ${sin}^{2} x + sin x = 0$ $sin^2x + sinx = 0$ from [0,2pi]?

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How do you solve the trig equation: ${sin}^{2} x + sin x = 0$ $sin^2x + sinx = 0$ from [0,2pi]?

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Answer 1 · 2015-04-16T11:52:26

You can write it, collecting $sin (x)$ $sin(x)$ , as:
$sin (x) [sin (x) + 1] = 0$ $sin(x)[sin(x)+1]=0$
So:
$sin (x) = 0$ $sin(x)=0$ for $x = 0, π, 2 π$ $x=0, pi, 2pi$
and
$sin (x) = - 1$ $sin(x)=-1$ for $x = \frac{3}{2} π$ $x=3/2pi$

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How do you solve the trig equation: ${sin}^{2} x + sin x = 0$ $sin^2x + sinx = 0$ from [0,2pi]?

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How do you solve the trig equation: ${sin}^{2} x + sin x = 0$ $sin^2x + sinx = 0$ from [0,2pi]?