How to solve the inequality sin2x>sin4x, if 0=

First, we'll have to move sin 4x to the left
side:

sin 2x - sin 4x >
0

The difference of the matching trigonometric functions
returns the product:

sin 2x - sin 4x = 2
sin[(2x-4x)/2]*cos[(2x+4x)/2]

sin 2x - sin 4x = 2
sin(-x)*cos(3x)

Since the sine function is odd, the
function sin(-x) = - sin x

sin 2x - sin 4x =
0

We'll re-write the
equation:

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

We'll cancel each
factor:

sin x = 0

x =
(-1)^k*arcsin 0 + k*pi

x =
k*pi

cos(3x) = 0

3x =
+/-arccos 0 + 2k*pi

3x = +/- (pi/2) +
2k*pi

x = +/- (pi/6) +
2k*pi/3

Now, we'll see where the functions sin x and cos 3x
are both positive, or both negative, for the product to be strictly
positive.

The functions sine and cosine are both positive
in the 1st quadrant, and they are both negative in the 3rd
quadrant.

sin x = 0 => x = 0 (1st
quadrant)

sin x = 0 => x = pi (3rd
quadrant)

cos 3x = 0 => 3x = pi/2 => x = pi/6
(1st quadrant)

cos 3x = 0 => 3x = pi + pi/2
=> x = pi/3 + pi/6

x = 3pi/6 => x = 3pi/2
(3rd quadrant)

For the inequality to hold,
the values of x belong to the reunion of intervals
(0;pi/6)U(pi/2;5pi/6)U(3pi/2;11pi/6)