#sin(pi/12)=sin(pi/3-pi/4)#
#=sin(pi/3)cos(pi/4)-cos(pi/3)sin(pi/4)#
#=sqrt3/2xx1/sqrt2-1/2xx1/sqrt2#
#=(sqrt3-1)/(2sqrt2)#
Again
#cos(pi/12)=cos(pi/3-pi/4)#
#=cos(pi/3)cos(pi/4)+sin(pi/3)sin(pi/4)#
#=1/2xx1/sqrt2+sqrt3/2xx1/sqrt2#
#=(sqrt3+1)/(2sqrt2)#
Now the given equation is
#(sqrt3-1)sintheta+(sqrt3+1)costheta=2#
Dividing both sides by #2sqrt2# we get
#(sqrt3-1)/(2sqrt2)sintheta+(sqrt3+1)/(2sqrt2)costheta=2/(2sqrt2)#
#=>sin(pi/12)sintheta+cos(pi/12)costheta=1/sqrt2#
#=>cos(theta-pi/12)=cos(pi/4)#
#=>theta-pi/12=2npipmpi/4" where "n in ZZ#
So
#=>theta=2npi+pi/4+pi/12" where "n in ZZ#
#=>theta=2npi+pi/3" where "n in ZZ#
Again
#theta=2npi-pi/4+pi/12" where "n in ZZ#
#=>theta=2npi-pi/6" where "n in ZZ#