I've worked this out two different ways but I think this third way is best. There are several double angle formulas for cosine. Let's not get tempted by any of them. Let's avoid squaring equations as well.
#cos 2x + 2 sin 2x + 2 = 0#
#cos 2x + 2 sin 2x = -2#
The linear combination of cosine and sine is a phase shifted cosine.
Let #r=\sqrt{1^2 + 2^2}# and
# theta = text{Arc}text{tan}(2/1) #
I indicated the principal inverse tangent, here in the first quadrant, around #theta=63.4^circ#. We're assured
#r cos theta = \sqrt{5} (1/\sqrt{5} ) = 1 #
# r sin theta = \sqrt{5} (2/\sqrt{5}) = 2 #
So we can rewrite our equation
#sqrt{5} ( (1/\sqrt{5} ) cos 2x + (2/\sqrt{5}) sin 2x) = -2#
# (1/\sqrt{5} ) cos 2x + (2/\sqrt{5}) sin 2x = -2/sqrt{5} #
# cos 2x cos theta + sin 2x sin theta = -2/sqrt{5} #
#cos(2x - theta) = sin(-theta)#
#cos(2x - theta) = cos(90^circ + theta)#
Always remember the general solution to #cos x = cos a# is #x= \pm a + 360^circ k quad # for integer #k#.
# 2x - theta = \pm (90^circ + theta) + 360^circ k #
#2x = theta \pm (90^circ + theta) + 360^circ k#
# x = theta/2 \pm (45^circ + theta/2) + 180^circ k#
Taking the signs one at a time,
# x = theta + 45^circ + 180^circ k or x = -45^circ + 180^circ k#
#phi = theta+45^circ# is a constant we can try to get a better expression for:
#tan(phi) = tan ( arctan(2)+45^circ ) #
# = { tan arctan(2) + tan (45^circ) }/{ 1- tan(arctan(2))tan(45^circ)} = {2 + 1}/{1 - 2} = -3#
We know #phi# is in the second quadrant, not in the usual range of the principal value.
#phi = text{Arc}text{tan}(-3) + 180^circ #
That turns out not to matter because we're adding #180^circ k# to #phi# in the general solution anyway. Putting it all together,
# x = arctan(-3) + 180^circ k or x = -45^circ + 180^circ k#
We don't have to be meticulous about the principal value of the arctan; since we're adding #180^circ k# any value will do. We could write the first #x=arctan(-3)# with the #180^circ k# implied, but let's leave it here.
