Question #dd6b9

1 Answer
Nov 27, 2017

#Sin2x=2^(1/2)cosx#

#=>2Sinxcosx-2^(1/2)cosx=0#

#=>2^(1/2)cosx(2^(1/2)sinx-1)=0#

Assuming #-360<=x<=360^@#

When #cosx=0#

we get greatest negative root of x.
#cosx=0=cos(-270^@)#

#=>x=-270^@#

When #2^(1/2)sinx-1=0#

we get lowest positive root of x.
#sinx=1/sqrt2=sin(45^@)#

#=>x=45^@#

So sum of these roots is #=-270^@+45^@=-225^@#