Question #6527d

#sin3x+cos3x=0#
#rArrcos(pi/2-3x)+cos3x=0#
#rArr2cos((pi/2-3x+3x)/2)cos((pi/2-3x-3x)/2)=0#
#rArr2cos(pi/4)cos(pi/4-3x)=0#
#rArrcos(pi/4-3x)=0,where, 2cos(pi/4)!=0#
#rArrcos(3x-pi/4)=0 #, as #cos(-theta)=costheta#
#rArr3x-pi/4=(2k+1)pi/2,kinZ,#
#rArr3x=(2k+1)pi/2+pi/4#
#color(red)(rArrx=(2k+1)pi/6+pi/12,kinZ)#,
So, taking # k=0,+-1.+-2,+-3...#
#k=0rArrx=pi/6+pi/12=color(blue)(pi/4in(0.pi),#
#k=1rArrx=(3pi)/6+pi/12=color(blue)((7pi)/12in(0.pi))#
# k=2rArrx=(5pi)/6+pi/12=color(blue)((11pi)/12in(0.pi),#
# k=3rArrx=(7pi)/6+pi/12=(15pi)/12!in(0.pi)#
Also,#k=-1rArrx=(-pi)/6+pi/12=-pi/12!in(0.pi)#
Hence, #x=pi/4,(7pi)/12,(11pi)/12#

Here,

#color(blue)(sin3x+cos3x=0) ,where,color(red)(x in (0,pi)#

Using above #color(blue)((I)and (II)#

#color(blue)(3sinx-4sin^3x+4cos^3x-3cosx=0#

#=>4cos^3x-4sin^3x-3cosx+3sinx=0#

#=>4(cos^3x-sin^3x)-3(cosx-sinx)=0#

#=>4(cosx-sinx)(cos^2x+cosxsinx+sin^2x)-3(cosx-sinx)=0#

#=>(cosx-sinx)[4(cos^2x+cosxsinx+sin^2x)-3]=0#

#=>(cosx-sinx)[4(1+cosxsinx)-3]=0#

#=>(cosx-sinx)[4+4sinxcosx-3]=0#

#=>(cosx-sinx)(1+2*2sinxcosx)=0#

#=>(cosx-sinx)(1+2color(violet)(sin2x))=0...tocolor(violet)(Apply(III)#

#=>cosx-sinx=0 or 1+2sin2x=0=>sin2x=-1/2#

Now, #x in (0,pi)=>I^(st)Quadrant or II^(nd)Quadrant#

#(i)cosx-sinx=0=>sinx=cosx=>tanx=1 > 0#

#i.e.I^(st)Quadrant=>color(red)(x=pi/4#

Also, #x in(0,pi) #

#=>2x in(0,2pi)=>I^(st),II^(nd),III ^(rd),IV^(th)Quadrant#

#(ii)sin2x=-1/2 < 0=>III^(rd) or IV^(th) Quadrant#

#:.2x=pi+pi/6=(7pi)/6 or 2x=2pi-pi/6=(11pi)/6#

#i.e. color(red)( x=(7pi)/12 or x=(11pi)/12#

Hence, #x=pi/4,(7pi)/12,(11pi)/12#