How do you find all solutions to the equation in the interval [0,2pi] given #sin3x = sinx#?

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Answer 1 · 2016-07-23T05:54:26

Solution set is #{0,pi/4,(3pi)/4,pi,(5pi)/4,(7pi)/4,2pi}#

As #sin3x=3sinx-4sin^3x#, the equation #sin3x=sinx# is nothing but

#3sinx-4sin^3x=sinx# or

#2sinx-4sin^3x=0# or

#sinx-2sin^3x=0# or #sinx(1-2sin^2x)=0#

Hence either #sinx=0# and #x=0, pi# or #2pi#

or #sin^2x=1/2# i.e. #sinx=+-1/sqrt2# and #x=pi/4,(3pi)/4,(5pi)/4# and #(7pi)/4#

Hence solution set is #{0,pi/4,(3pi)/4,pi,(5pi)/4,(7pi)/4,2pi}#