Solve the equation #sin2v=sin^2v#?

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Answer 1 · 2017-07-25T16:51:50

#v=npi# or #npi+1.107#, where angle is in radians and #n# is an integer.

As #sin2v=sin^2v#

we have #2sinvcosv-sin^2v=0#

or #sinv(2cosv-sinv)=0#

i.e. either #sinv=0# i.e. #v=npi#, where #n# is an integer

or #2cosv-sinv=0# i.e. #tanv=2# and #v=npi+-tan^(-1)2#

or #v=npi+-tan^(-1)2=npi+1.107# (in radians)

Answer 2 · 2017-07-25T16:52:03

Given

#sin^2v=sin2v#

#=>sin^2v-sin2v=0#

#=>sin^2v-2sinvcosv=0#

#=>sinv(sinv-2cosv)=0#

when

#sinv=0#

#=>v=npi" where "n in ZZ#

When

#(sinv-2cosv)=0#

#=>sinv=2cosv#

#=>tanv=2=tanalpha#, where #tan^-1 2=alpha#

#=>v=npi+alpha" where " n in ZZ#