sin 2 theta2 3 cos 2 theta2SolutionI assume this is a trigo

(sin 2 theta)^2 = (3 cos 2 theta)^2

I assume this is a trigonometric equation we have to solve :

(sin2x)^2 = (3cos2x)^2 in interval [ 0, 2pi)

(sin2x)^2 - (3cos2x)^2 =0

we know cos^2x = 1- sin^2x

So, (sin2x)^2 - 3(1- sin^2x ) =0

sin^2(2x) -3 +3sin^2(2x) =0

5sin^2(2x) -3 =0

sin(2x) = +/- sqrt(3/5)

= +/- 0.77

2x = sin^-1(0.77) = 0.87, pi-0.87= 0.87, 2.26 rad

x = 0.435 rad , 1.13 rad

sin(2x) = -0.77

2x = pi+0.87 rad = 4.02 rad , 2pi -0.87 rad = 5.41 rad

x = 2.01 rad , 2.70 rad

Solution : x = 0.87 , 2.01, 2.26 , 2.70 radians in interval [ 0, 2pi)