How do you find the exact solutions of the equation $sin 2 x - sin x = 0$ $sin2x-sinx=0$ in the interval $[0, 2 π)$ $[0,2pi)$ ?

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How do you find the exact solutions of the equation $sin 2 x - sin x = 0$ $sin2x-sinx=0$ in the interval $[0, 2 π)$ $[0,2pi)$ ?

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Answer 1 · 2017-02-20T04:15:39

$0, \frac{π}{3}, π, \frac{5 π}{3}$ $0, pi/3, pi, (5pi)/3$ ,

Explanation:

Use trig identity: sin 2 x = 2sin x.cos x.
Replace in the equation sin 2x = 2sin x.cos x
2sin x.cos x - sin x = 0
sin x(2cos x - 1) = 0
a. sin x = 0 --> $x = 0, x = π, x = 2 π$ $x = 0, x = pi, x = 2pi$
b. 2cos x - 1 = 0 --> $cos x = \frac{1}{2}$ $cos x = 1/2$
Unit circle gives:
$x = \frac{1}{2}$ $x = 1/2$ --> $cos x = \pm \frac{π}{3}$ $cos x = +- pi/3$
Note. $- \frac{π}{3}$ $- pi/3$ and $\frac{5 π}{3}$ $(5pi)/3$ are co-terminal.
Answers for $(0, 2 π)$ $(0, 2pi)$
$0, \frac{π}{3}, π, \frac{5 π}{3}, 2 π$ $0, pi/3, pi, (5pi)/3, 2pi$

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How do you find the exact solutions of the equation $sin 2 x - sin x = 0$ $sin2x-sinx=0$ in the interval $[0, 2 π)$ $[0,2pi)$ ?

1 Answer

Explanation:

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How do you find the exact solutions of the equation $sin 2 x - sin x = 0$ $sin2x-sinx=0$ in the interval $[0, 2 π)$ $[0,2pi)$ ?