If #cos (x) cos (x - pi/3) = 1 - n/6, then the boundary value "n" that meets is = ?

1 Answer
Oct 12, 2017

#cos (x) cos ((x) - phi/3) = 1 - n/6#

#=>2cos (x) cos (x - pi/3) = 2 - n/3#

#=>cos (2x-pi/3) +(cos (pi/3) = 2 - n/3#

#=>cos (2x-pi/3) +1/2 = 2 - n/3#

#=>cos (2x-pi/3) = 3/2- n/3#

But we know

#-1<= costheta<=1#

#=>-1<= cos(2x-pi/3)<=1#

#=>-1<= 3/2-n/3<=1#

#=>-1-3/2<= 3/2-n/3-3/2<=1-3/2#

#=>-5/2<= -n/3<=-1/2#

#=>-5/2*(-3)>= -n/3*(-3)>=-1/2*(-3)#

#=>7.5>= n >=1.5#