How do you solve for the exact solutions in the interval [0,2pi] of sin 2x=sinx?

1 Answer
Jim G.
Mar 4, 2016

x = 0 , pi/3 ,(5pi)/3 ,2pi

Explanation:

Using appropriate color(blue)" Double angle formula "

• sin2A = 2sinAcosA

hence : 2sinxcosx = sinx

and 2sinxcosx - sinx = 0

Take out common factor of sinx

thus : sinx(2cosx - 1 ) = 0

and so sinx = 0 or cosx =1/2

sinx = 0 rArr x = 0 or 2pi

cosx = 1/2 rArr x = pi/3 or (2pi - pi/3 ) = (5pi)/3

These are the solutions in the interval [0 , 2pi]

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