How do you solve #sin(2x)cos(x)=sin(x)#?
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"If the momentum of an object increases by #20%#, what will be the percent increase in its kinetic energy?"
#x=npi,2npi+-(pi/4), and 2npi+-((3pi)/4)# where #n in ZZ#
#rarrsin2xcosx=sinx#
#rarr2sinx*cos^2x-sinx=0#
#rarrsinx(2cos^2x-1)=0#
#rarrrarrsinx*(sqrt2cosx+1)*(sqrt2cosx-1)=0#
When #sinx=0#
#rarrx=npi#
When #sqrt2cosx+1=0#
#rarrcosx=-1/sqrt2=cos((3pi)/4)#
#rarrx=2npi+-((3pi)/4)#
When #sqrt2cosx-1=0#
#rarrcosx=1/sqrt2=cos(pi/4)#
#rarrx=2npi+-(pi/4)#
#x = npi, pi/4 + npi, (3pi)/4 + npi# where #n in ZZ#
We have,
#color(white)(xxx)sin2xcosx = sinx#
#rArr 2sinxcosx xx cosx = sinx# [As, #sin 2x = 2sinxcosx#]
#rArr 2sinxcos^2x - sin x = 0#
#rArr sinx(2cos^2 - 1) = 0#
Now,
Either,
#sin x = 0 rArr x = sin^-1(0) = npi#, where #n in ZZ#
Or,
#color(white)(xxx)2cos^2x - 1 = 0#
#rArr 2cos^2x - (sin^2x + cos^2x) = 0# [As #sin^2x + cos^2 x = 1#]
#rArr 2cos^2x-sin^2x-cos^2x = 0#
#rArr cos^2x - sin^2x = 0#
#rArr (cosx + sin x)(cos x - sin x) = 0#
So, Either #cos x - sin x = 0 rArr cos x = sin x rArr x = pi/4 +- npi#, where #n in ZZ#
Or,
#cos x + sin x = 0 rArr cos x = -sinx rArr x = (3pi)/4 +- npi#, where #n in ZZ#
So, Summing it all up,
#x = npi, pi/4 +- npi, (3pi)/4 +- npi#, where #n in ZZ#
