Auxilliary angle method:
Let 3cos 2x - 2sin 2x = Rcos(2x + alpha), where R > 0 and alpha is an acute angle.
Expanding, we get: 3cos 2x - 2sin 2x = R(cos 2x * cos alpha - sin 2x * sin alpha)
Equating coefficients of cos 2x: 3 = Rcos alpha ______ (1)
Equating coefficients of sin 2x: -2 = -Rsin alpha ______ (2)
(1)^2 + (2)^2: 9 + 4 = R^2 * (cos^2 alpha + sin^2 alpha) = R^2
So, R = sqrt(13, as R > 0
(2) / (1): -Rsin alpha / Rcos alpha = -2 / 3
tan alpha = 2 / 3
alpha = 0.58002 ..., as alpha is acute
So, 3cos 2x - 2sin 2x = sqrt(13) * cos(2x + 0.58002...)
Thus, we can transform our original problem, 3cos 2x - 2sin 2x = 1, 0 <= x <= 2 * pi
into sqrt(13) * cos(2x + 0.58802...) = 1, 0 <= 2x <= 4 * pi
and so cos(2x + 0.58802...) = 1 / sqrt(13), 0.58802... <= 2x + 0.58802... <= 13.154...
Let beta be the related angle, acute, satisfying cos beta = 1 / sqrt(13)
Thus, beta = 1.2897...
Since cos(2x + 0.58802...) is positive, we need solutions in the first and fourth quadrants.
So, 2x + 0.58802... = beta, 2 * pi - beta, 2 * pi + beta, 4 * pi - beta in the desirred domain.
So, 2x + 0.58802... = 1.2897..., 4.9934..., 7.5729..., 11.276...
2x = 0.70175..., 4.4054..., 6.9849..., 10.688...
x = 0.35087..., 2.2027..., 3.4924..., 5.3443...
So, x = 0.3509, 2.202, 3.492, 5.344 (4 sig fig).