Given #2cosx+2sinx=sqrt(6)#
First divide both sides by 2 to get:
#cosx+sinx=sqrt(6)/2#
Squaring both sides we see:
#3/2 = 6/4 = (sqrt(6)/2)^2#
#= (cosx+sinx)^2 = cos^2x+2sinxcosx+sin^2x#
#= 1+2sinxcosx = 1+sin2x#
Subtracting 1 from both ends of this equation, we get:
#sin2x = 1/2#
Hence #2x = pi/6+2npi# or #2x = (5pi)/6+2npi# (#n# in #ZZ#)
Dividing both sides by #2# we find:
#x=pi/12+npi# or #x=(5pi)/12+npi# (#n# in #ZZ#)
For even values of #n#, these angles are in QI so #cos x > 0# and #sin x > 0# giving a solution of the original problem. For odd values of #n#, these angles are in QIII so #cos x < 0# and #sin x < 0# resulting in #2cosx+2sinx=-sqrt(6)#, so we have to reject odd values of #n#. This happens because we squared both sides of the equation.
So solutions of the original problem are of the form:
#x=pi/12+2npi# or #x=(5pi)/12+2npi# (#n# in #ZZ#)
