Solve the equation 3sin^2 2x= cos^2 2x for 0 <x <180?

0° < x < 180°
Then, 0° < 2x < 360°

3 sin²(2x) = cos²(2x)
sin²(2x)/cos²(2x) = 1/3
tan²(2x) = 1/3
tan(2x) = 1/√3 or tan(2x) = -1/√3
2x = 30°, (180°+30°) or 2x = (180°-30°), (360°-30°)
x = 15°, 75°, 105°, 165°

3sin^2 2x = cos^2 2x

3sin^2 2x / cos^2 2x = 1

3tan^2 2x = 1

tan^2 2x = 1/3

tan 2x = ± (√3)/3

2x = kpi ± pi/6

x = kpi/2 ± pi/12 , k any integer

for k = 0 and k = 1 and k = 2, answers will be on (0°, 180°) :

x = pi/12, 5pi/12, 7pi/12, 11pi/12

Answers on angles mode :

x = 15°, 75°, 105° and 165°

Alternative method

3sin²(2x) = 1 - sin²(2x)

4sin²(2x) = 1

sin²(2x) = 1/4

sinx = ±1/2

2x = 30°, 150°, 210°, 330°

x = 15°, 75°, 105°, 165°

4 sin² (2x) = 1
sin² (2x) = 1 / 4
sin (2x) = ± 1 / 2
2x = 30⁰ , 150⁰ , 390⁰ , 510⁰
x = 15⁰ , 75⁰ , 195⁰ , 255⁰

x = 15⁰ , 75⁰_____for given range of x

sin(2x)^2 / cos(2x)^2 = 1/3
tan(2x)^2 = 3/9
tan(2x) = +/- sqrt(3) / 3
2x = 30 + 180 * k , -30 + 180 * k
x = 15 + 90k , -15 + 90k
x = 15 , 105 , 75 , 165