Question #6a844

1 Answer
Feb 25, 2017

x=(3pi)/2,pi/10,pi/2" when "x in [0.2pi]x=3π2,π10,π2 when x[0.2π]

Explanation:

sin2x-cos3x=0sin2xcos3x=0

=>sin2x-sin(pi/2-3x)=0sin2xsin(π23x)=0

=>2cos(1/2(2x+pi/2-3x))sin(1/2(2x-pi/2+3x))=02cos(12(2x+π23x))sin(12(2xπ2+3x))=0

=>cos(x/2-pi/4)sin((5x)/2-pi/4)=0cos(x2π4)sin(5x2π4)=0

when

cos(x/2-pi/4)=0=cos(pi/2)=cos ((3pi)/2)cos(x2π4)=0=cos(π2)=cos(3π2)

cos(x/2-pi/4)=0=cos(pi/2)cos(x2π4)=0=cos(π2)

=>cos(x/2-pi/4)=cos(pi/2)cos(x2π4)=cos(π2)

=>x/2=pi/2+pi/4=(3pi)/4x2=π2+π4=3π4

=>x=(3pi)/2x=3π2

Again

cos(x/2-pi/4)=cos ((3pi)/2)cos(x2π4)=cos(3π2)

=>x/2-pi/4=(3pi)/2x2π4=3π2

=>x=3pi+pi/2>2picolor(red)" not wanted"x=3π+π2>2π not wanted

when

sin((5x)/2-pi/4)=0=sin0=sin(pi)sin(5x2π4)=0=sin0=sin(π)

=>(5x)/2-pi/4=05x2π4=0

=>x=pi/4xx2/5=pi/10x=π4×25=π10

Again

sin((5x)/2-pi/4)=sin(pi)sin(5x2π4)=sin(π)

=>(5x)/2=pi+pi/4=(5pi)/45x2=π+π4=5π4

=>x=pi/2x=π2