What is the solution of the equation 2six*sin3x-cos4x=0?

We'll shift the term cos 4x to the right
side:

2 sin x*sin 3x = cos
4x

Now, we'll transform the left side into a difference of
2 cosines.

2 sin x*sin 3x = cos b - cos
a

We'll consider x = (a+b)/2 => a + b = 2x
(1)

3x = (b-a)/2 => -a + b = 6x
(2)

We'll add (1)+(2):

a + b -
a + b = 2x + 6x

2b = 8x

b =
4x

a = -2x

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

The equation will
become:

cos(4x) - cos (-2x) = cos
4x

We'll reduce like terms:

-
cos (-2x) = 0

The function cosine is even, therefore cos
(-2x) = cos (2x)

cos (2x) =
0

2x = +/- pi/2 + 2k*pi

x =
+/- pi/4 + 2k*pi

The solutions of the
equation are represented by the set {+/- pi/4 +
2k*pi}.