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