Given: #-sqrt(3) cos x + sin x = - sqrt(3)#
Rearrange: #" "sqrt(3) - sqrt(3) cos x + sin x = 0#
When we think of #sqrt(3)# in terms of cosine, realize #cos(pi/6) = sqrt(3)/2#
Group the second two terms and multiply and divide both terms by #2/2#:
#sqrt(3) - 2((sqrt(3))/2 cos x - 1/2sin x) = 0#
Substitute in #cos (pi/6) " for " sqrt(3)/2 " and " sin (pi/6) " for " 1/2:#
#sqrt(3) - 2(cos (pi/6) cos x - sin (pi/6) * sin x) = 0#
Realize that #" "cos(x + pi/6) = cos (pi/6) cos x - sin (pi/6) * sin x#
#sqrt(3) - 2cos(x + pi/6) = 0#
#- 2cos(x + pi/6) = -sqrt(3)#
#cos(x + pi/6) = sqrt(3)/2#
Take the inverse cosine of both sides:
#cos^-1(cos(x + pi/6)) = cos^-1(sqrt(3)/2)#
#x + pi/6 = 2 pi n + pi/6#
and #x + pi/6 = 2 pi n + (11 pi)/6#
Simplify:
#x = 2 pi n " and " x = 2 pi n +(10 pi/6) = 2 pi n + (5 pi)/3#