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# How do you solve the equation sin2x-sinx=0 on the interval [0,2π]?

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Gió Share
Feb 17, 2015

You can start by writing:
$\sin \left(2 x\right) = 2 \sin \left(x\right) \cos \left(x\right)$
So you get:
$2 \sin \left(x\right) \cos \left(x\right) - \sin \left(x\right) = 0$
$\sin \left(x\right) \cdot \left[2 \cos \left(x\right) - 1\right] = 0$

This holds when:

$\sin \left(x\right) = 0$ and then $x = 0 \mathmr{and} x = \pi \mathmr{and} x = 2 \pi$

or when:

$2 \cos \left(x\right) - 1 = 0$
$\cos \left(x\right) = \frac{1}{2}$ and then $x = \frac{\pi}{3} \mathmr{and} x = \frac{5}{3} \pi$

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