How do you solve $2sin^2x + sinx =0$ from $[0,2pi]$ ?

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Answer 1 · 2015-07-25T15:03:15

Solve $2sin^2 x + sin x = 0$ $[0, 2pi]$

$sin x(2sin x + 1) = 0$
Use trig table and unit circle -->

a. sin x = 0 => x = 0, and $x = pi$ and $x = 2pi$

b. 2sin x + 1 = 0 => $sin x = -1/2$ => $x = (7pi)/6$ and $x = (11pi)/6$ .

Finally, within interval $[0, 2pi],$ there are 5 answers:
$0, pi, 2pi, (7pi)/6 and (11pi)/6$

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