Calculate (sin x)^2 + sin (2x)=0?
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(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}
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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
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}
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