How do you solve #sin^2x - 2sinx = 0#?

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no idea

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Answer 1 · 2018-04-10T05:22:28

#x=npi# where #n# is an integer.

As #sin^2x-2sinx=0#, we have

#sinx(sinx-2)=0#

As #sinx-2!=0#, we must have #sinx=0#

i.e. #x=npi# where #n# is an integer.