Find the solution of the equation sin^2 2x - 2sin 4x+3cos^2 2x = 0

First, we'll use the double angle identity to re-write the
middle term.

sin 4x = sin 2(2x) = 2sin 2x*cos
2x

We'll re-write the
equation:

(sin 2x)^2 - 4 sin 2x*cos 2x + 3(cos 2x)^2 =
0

We'll divide by (cos 2x)^2 the
equation:

(tan 2x)^2 - 4 tan 2x + 3 =
0

We'll substitute tan 2x by
t.

We'll re-write the
equation:

t^2 - 4t + 3 =
0

We'll apply quadratic
formula:

t1 = [4+sqrt(16 -
12)]/2

t1 = (4+2)/2

t1 =
3

t2 = 1

We'll put tan 2x = t1
=> tan 2x = 1 => 2x = arctan1 + k*pi

2x =
pi/4 + k*pi

We'll divide by
2:

x = pi/8 + k*pi/2

Let tan
2x = t2 => tan 2x = 3 => 2x = arctan3 +
k*pi

x = [arctan(3)]/2 +
k*pi/2

The requested solutions of the
equation are: {pi/8 + k*pi/2 / k integer number}U{[arctan(3)]/2 + k*pi/2 / k integer
number}.