What is x if (1+cos2x)/2=sin(-2x)?

We'll recognize in the first term the formula of the half
angle.

(1+cos2x)/2 = (cos
x)^2

We also know that the sine function is odd, so
sin(-2x) = - sin 2x.

We'll re-write the equation,
substituting sin 2x by 2sinx*cosx.

We'll re-write now the
entire expression.

(cos x)^2 = - 2sin x * cos
x

We'll add 2sin x * cos x both
sides:

(cos x)^2 + 2sin x * cos x =
0

We'll factorize by cos x and we'll
get:

cos x * (cos x + 2sin x) =
0

We'll put each factor from the product as
0.

cos x = 0

This is an
elementary equation.

x = arccos 0 +
2k*pi

x = pi/2 +
2k*pi

or

x = 3pi/2 +
2k*pi

cos x + 2sin x = 0

This
is a homogeneous equation, in sin x and cos x.

We'll divide
the entire equation, by cos x.

1 + 2 sinx/cos x =
0

But the ratio sin x / cos x = tg x. We'll substitute the
ratiosin x / cos x by tg x.

1 + 2tan x=
0

We'll subtract 1 both
sides:

2tan x = -1

We'll
divide by 2:

tan x = -1/2

x =
arctg(-1/2 ) +k*pi

x = - arctg(1/2) +
k*pi

The solutions of the equation
are:

{pi/2 + 2k*pi}U{3pi/2 + 2k*pi}U{-
arctg(1/2) + k*pi}