#II^(nd) method :#
We know that,
#color(brown)((1)sintheta=cos(pi/2-theta)#
#color(red)((2)cosC-cosD=-2sin((c+D)/2)sin((C-D)/2)#
#color(blue)((3)sintheta=0=>theta=kpi ,k inZZ#
Here,
#cos(pi/4-x)=color(brown)(sin(pi/6+x)...tocolor(brown)(Apply(1)#
#=>cos(pi/4-x)=color(brown)(cos[pi/2-(pi/6+x)]#
#=>cos(pi/4-x)=cos(pi/2-pi/6-x)#
#=>cos(pi/4-x)=cos(pi/3-x)#
#=>color(red)(cos(pi/4-x)-cos(pi/3-x)=0#
Using #color(red)((1)#,we get
#=>color(red)(-2sin((pi/4-x+pi/3-x)/2)sin((pi/4-x-pi/3+x)/2)=0#
#=>-2sin(((7pi)/12-2x)/2)sin((-pi/12)/2)=0#
#=>sin((7pi)/24-x)=0to[because-2sin(-pi/24)!=0]#
#=>color(blue)(sin(x-(7pi)/24)=0...tocolor(blue)(Apply(3)#
#=>color(blue)(x-(7pi)/24=kpi, k inZZ#
#=>x=kpi+(7pi)/24 ,k inZZ#