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)
