How do you solve the equation $sin^2theta-sintheta=0$ for $[0,2pi)$ ?

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Answer 1 · 2016-12-02T15:21:39

$0, pi/2, pi, 2pi$

$sin^2 t - sin t = 0$
$sin t(sin t - 1) = 0$
Use unit circle -->
a. sin t = 0 --> 3 solution arcs -->
arc t = 0, arc $t = pi$ , and arc $t = 2pi$

b. sin t - 1 = 0 --> sin t = 1 --> arc $t = pi/2$

Answers for $(0, 2pi)$
$0, pi/2, pi, 2pi$

How do you solve the equation sin^2theta-sintheta=0sin^2theta-sintheta=0 for [0,2pi)[0,2pi)?