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

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We have 2*sin x*sin 3x - cos 4x = 0

(sin a)(sin b) = (1/2)(cos (a - b) - cos (a + b))

=> 2*(sin a)(sin b) = cos (a - b) - cos (a + b)

2*sin x*sin 3x - cos 4x = 0

=> cos (x - 3x) - cos 4x - cos 4x = 0

=> cos -2x = 0

=> -2x = 90 and -2x = 270

=> x = -45 and x = -135

**The solution of the equation is (-45 + n*360, -135+n*360)**

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}.**