To determine the solutions of 1 - cos 2x = 4*sin x, first convert cos 2x to a form that has only terms with sin x

1 - cos 2x = 4*sin x

=> 1 - (1 - 2*(sin x)^2) = 4*sin x

=> 1 - 1 + 2*(sin x)^2 = 4*sin x

=> (sin x)^2 = 2*sin x

=> (sin x)(sin x - 2) = 0

sin x = 0

=> x = n*360 and 180 + n*360

sin x - 2 = 0

=> sin x = 2

which is not possible

**The solutions of the equation are (n*360, 180 + n*360)**

We'll apply the following formula:

1 - cos 2x = 2*[sin (2x/2)]^2

1 - cos 2x = 2*(sin x)^2 (1)

We'll re-write the equation, replacing the left side by (1):

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

We'll divide by 2:

(sin x)^2 = 2*sin x

We'll move all terms to on side:

(sin x)^2 - 2*sin x = 0

We'll factorize by sin x:

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

We'll cancel each factor:

sin x = 0

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

x = k*pi

sin x = 2 impossible since the value of sine function cannot be larger than 1.

**The only possible set of solutions ****of the trigonometric equation,****, for any ****integer ****k, **** is: {k*pi, }.**