Determine x if sin2x=-(sinx)^2.

The first step is to re-write the expression, using the
formula for the double angle:

sin 2x =
2sinx*cosx.

We'll re-write now the entire expression,
moving all terms to one side.

(sin x)^2+2sin x * cos x =
0

We'll factorize and we'll
get:

sin x * (sin x + 2cos x) =
0

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

sin x = 0

We notice that
it is an elementary equation.

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

x = k*pi

We'll put the
next factor as zero.

sin x + 2cos x =
0

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

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

sin x / cos x + 2 = 0

But
the ratio sin x / cos x = tg x.

We'll substitute the
ratiosin x / cos x by tg x.

tg x + 2 =
0

x = arctg(-2 ) +k*pi

x = pi
- arctg2 + k*pi

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

The x values for the expression to be true
are:

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