Determine all solutions of the trigonometric equation 1-cos2x=4sinx.

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,
}.