sin2x-cos3x=0sin2x−cos3x=0
=>sin2x-sin(pi/2-3x)=0⇒sin2x−sin(π2−3x)=0
=>2cos(1/2(2x+pi/2-3x))sin(1/2(2x-pi/2+3x))=0⇒2cos(12(2x+π2−3x))sin(12(2x−π2+3x))=0
=>cos(x/2-pi/4)sin((5x)/2-pi/4)=0⇒cos(x2−π4)sin(5x2−π4)=0
when
cos(x/2-pi/4)=0=cos(pi/2)=cos ((3pi)/2)cos(x2−π4)=0=cos(π2)=cos(3π2)
cos(x/2-pi/4)=0=cos(pi/2)cos(x2−π4)=0=cos(π2)
=>cos(x/2-pi/4)=cos(pi/2)⇒cos(x2−π4)=cos(π2)
=>x/2=pi/2+pi/4=(3pi)/4⇒x2=π2+π4=3π4
=>x=(3pi)/2⇒x=3π2
Again
cos(x/2-pi/4)=cos ((3pi)/2)cos(x2−π4)=cos(3π2)
=>x/2-pi/4=(3pi)/2⇒x2−π4=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(5x2−π4)=0=sin0=sin(π)
=>(5x)/2-pi/4=0⇒5x2−π4=0
=>x=pi/4xx2/5=pi/10⇒x=π4×25=π10
Again
sin((5x)/2-pi/4)=sin(pi)sin(5x2−π4)=sin(π)
=>(5x)/2=pi+pi/4=(5pi)/4⇒5x2=π+π4=5π4
=>x=pi/2⇒x=π2