(sinx)^2+sin2x=0

We know that sin2x= 2sinx*cosx

==> (sinx)^2+2sinx*cosx=0

==> sinx(sinx+2cosx)=0

Then sinx=0 or sinx+2cosx=0

First sinx=0

==> x= n*pi , (n=0,1,2,...)

For the other part:

sinx+2xosx=0

==> sinx=-2cosx

==> sinx/cosx= -2

==> tanx=-2

==> x= (n+1)pi-arctan 2

Theb x values are: {n*pi, (n+1)*pi-arctan 2}

To calculate the expression means to find out the x values for the equation to hold.

For the beginning, we'll re-write sin 2x = 2sinx*cosx.

We'll re-write now the entire expression.

(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

This is an elementary equation.

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

x = k*pi

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 give to hold are:

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

The roots of `sin^2x + sin(2x)=0` are required.

Use the formula for sin 2x = 2*sin x*cos x, this gives:

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

Factor sin x from the experssion

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

For sin x = 0, `x = sin^-1(0) = n*2*pi` and `pi + n*2*pi`

For sin x + 2*cos x = 0

sin x = -2*cos x

tan x = -2

x = `tan^-1(-2) +n*pi`

The roots of the given equation are `n*2*pi` , `pi + n*2*pi` and `tan^-1(-2) +n*pi`

To solve sin^2x+sin2x = 0

Solution:

We know that sin2x = 2sinx cosx.Replacing this in the given equation, we get:

sin^2x +2sinxcosx= 0. Or

sinx(sinx +2cosx ) = 0. Or

sinx = 0 or x = npi, n = 0,1,2,'..... Or

sinx +2cosx = 0 Or

sinx/cosx = - 2. Or tanx = -2. Or x = arc tan(-2) = npi-arc tan 2