Question #6a844

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Answer 1 · 2017-02-25T03:05:30

#x=(3pi)/2,pi/10,pi/2" when "x in [0.2pi]#

#sin2x-cos3x=0#

#=>sin2x-sin(pi/2-3x)=0#

#=>2cos(1/2(2x+pi/2-3x))sin(1/2(2x-pi/2+3x))=0#

#=>cos(x/2-pi/4)sin((5x)/2-pi/4)=0#

when

#cos(x/2-pi/4)=0=cos(pi/2)=cos ((3pi)/2)#

#cos(x/2-pi/4)=0=cos(pi/2)#

#=>cos(x/2-pi/4)=cos(pi/2)#

#=>x/2=pi/2+pi/4=(3pi)/4#

#=>x=(3pi)/2#

Again

#cos(x/2-pi/4)=cos ((3pi)/2)#

#=>x/2-pi/4=(3pi)/2#

#=>x=3pi+pi/2>2picolor(red)" not wanted"#

when

#sin((5x)/2-pi/4)=0=sin0=sin(pi)#

#=>(5x)/2-pi/4=0#

#=>x=pi/4xx2/5=pi/10#

Again

#sin((5x)/2-pi/4)=sin(pi)#

#=>(5x)/2=pi+pi/4=(5pi)/4#

#=>x=pi/2#