What is the general solution for sin^2(2x)=3cos^2(2x) ?

The general solution for sin^2(2x)=3cos^2(2x) is x= pi/3+(pin)/2 ,x= pi/6+(pin)/2

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

sin^{2} (2 x) = 3 cos^{2} (2 x)

x = \frac{π}{3} + \frac{πn}{2}, x = \frac{π}{6} + \frac{πn}{2}

Solution steps

sin^{2} (2 x) = 3 cos^{2} (2 x)

Subtract

3 cos^{2} (2 x)

from both sides

sin^{2} (2 x) - 3 cos^{2} (2 x) = 0

Factor

sin^{2} (2 x) - 3 cos^{2} (2 x) : (sin (2 x) + 3 cos (2 x)) (sin (2 x) - 3 cos (2 x))

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

Solving each part separately

sin (2 x) + 3 cos (2 x) = 0 or sin (2 x) - 3 cos (2 x) = 0

sin (2 x) + 3 cos (2 x) = 0 : x = \frac{π}{3} + \frac{πn}{2}

sin (2 x) - 3 cos (2 x) = 0 : x = \frac{π}{6} + \frac{πn}{2}

Combine all the solutions

x = \frac{π}{3} + \frac{πn}{2}, x = \frac{π}{6} + \frac{πn}{2}

- 2 sin (θ) = 2 sin (2 θ)

cos (θ) = \frac{π}{2}

2 sin (4 x) + 3 = 0

sin (α) = 0

sin (9 0^{\circ}) = cos (x)

What is the general solution for sin^2(2x)=3cos^2(2x) ?
The general solution for sin^2(2x)=3cos^2(2x) is x= pi/3+(pin)/2 ,x= pi/6+(pin)/2