Trig/Calculus help please

(9 sin (2x) + 9 cos (2x))^2 = 3

solve for the exact solution(s) of x, [0, 2pi)

9^2 * (sin(2x) + cos(2x))^2 = 3
81 * (sin(2x)^2 + 2sin(2x)cos(2x) + cos(2x)^2) = 3
81 * (1 + sin(2 * 2x)) = 3
81 * (1 + sin(4x)) = 3
81 + 81 * sin(4x) = 3
27 + 27 * sin(4x) = 1
27 * sin(4x) = -26
sin(4x) = -26/27
4x = arcsin(-26/27)
4x = 4.985401602709300151182251499983 + 2 * pi * k , 4.4393763580600795642056786498555 + 2 * pi * k

4x = 4.9854 + 2pi * k , 4.4394 * 2pi * k
x = 1.24635 + (pi/2) * k , 1.10985 + (pi/2) * k

x = 1.10985 , 1.24635 , 2.68065 , 4.25144 , 5.82224, 2.81715 , 4.38794 , 5.95874

(9 sin (2x) + 9 cos (2x))^2 = 3

81( sin2x + cos2x)^2= 3

(sin2x+ cos2x)^2 = 1/27

sin^2 (2x) + 2sin2xcos2x + cos^2(2x) = 1/27
Note: Sin^2(2x) + cos^2(2x)= 1 and 2sin2xcos2x = sin4x

Sin(4x) = 1/27-1

Sin(4x)= -26/27

Sine is negative in quadrants Iii and IV and sin^-1( -26/27) = 1.29778..., rounding to 1.3

4x = pi+ 1.3 in quadrant III, and 2pi-1.3 in quadrant IV

But with a multiple angle, continue around for 4 periods : add 2pi to the answers

4x= (pi+ 1.3) + 2pi* n or (2pi-1.3) + 2pi*n

4x = 4.442+ 6.28n or 4.983 + 6.28n

I'm not sure what they mean by exact answers, because these are decimal approximations, not special angles.

Dividing by 4:
X= {1.11+ 1.57n } or { 1.25+ 1.57n}

X= { 1.11, 2.68, 4.25, 5.8} or { 1.25, 2.82, 4.39, 5.96}

(These are all between 0 and 6.28)

I hope this helps!

Simplifying it
Sin(2x) + Cos(2x) = 1/(3√3)
√2 ((1/√2) Sin (2x) + (1/√2) Cos(2x)) = 1/(3√3)
Cos (π/4) Sin(2x) + Sin(π/4) Cos(2x) = 1/(3√6)
Sin((π/4) + 2x) = 1/(3√6)
π/4 + 2x = Sin⁻¹(1/(3√6))
x = (1/2) Sin⁻¹(1/(3√6)) - π/8