How do you solve $sin 2 x = sin x$ $sin 2x=sinx$ over the interval 0 to 2pi?

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Answer 1 · 2016-03-02T11:23:47

Solution is $x = {0, \frac{π}{3}, π, \frac{5 π}{3}, 2 π}$ $x={0, pi/3, pi, (5pi)/3, 2pi}$

As $sin 2 x = 2 sin x cos x$ $sin2x=2sinxcosx$ ,

$sin 2 x = sin x$ $sin2x=sinx$ is equivalent to

$2 sin x cos x = sin x$ $2sinxcosx=sinx$ or $2 sin x cos x - sin x = 0$ $2sinxcosx-sinx=0$ or

$sin x (2 cos x - 1) = 0$ $sinx(2cosx-1)=0$ i.e.

either $sin x = 0$ $sinx=0$ and $x = 0, π, 2 π$ $x=0, pi, 2pi$

or $2 cos x - 1 = 0$ $2cosx-1=0$ i.e. $cos x = \frac{1}{2}$ $cosx=1/2$ i.e. $x = \frac{π}{3}, \frac{5 π}{3}$ $x=pi/3, (5pi)/3$

Hence, Solution is $x = {0, \frac{π}{3}, π, \frac{5 π}{3}, 2 π}$ $x={0, pi/3, pi, (5pi)/3, 2pi}$

How do you solve sin 2x=sinx sin2x=sinxsin 2x=sinx over the interval 0 to 2pi?