How do you solve #sinxtan^2x=sinx#?

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Answer 1 · 2018-05-11T04:42:55

#x=npi# or #x=npi+-pi/4# where #n# is an integer

As #sinxtan^2x=sinx#

#sinxtan^2x-sinx=0#

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

i.e. either #sinx=0=sin0# i.e. #x=npi#

or #tan^2x=1=tan^2(pi/4)#

i.e. #tanx=+-tan(pi/4)#or #x=npi+-pi/4#

where #n# is an integer