(cos x)^2 + sin 2x = 0
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(cosx)^2 + sin2x = 0
We know that:
sin2x = 2sinx*cosx
==> cos^2 x + 2sinx*cosx = 0
Facros cos(x):
==> cosx*(cosx + 2sinx) = 0
Then :
cosx = 0 OR cosx + 2sinx = 0
cosx = 0 ==> x=pi/2 , 3pi/2
cosx + 2sinx = 0
==> cosx = -2sinx
==> -cosx/2 = sinx
Divide by cosx
==> -1/2 = tanx
==> tanx = -1/2
==> x= -arcttan (1/2)
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To solve cos^2 x+sin2x = 0
We know sin2x = 2sinx cox.
So the given equation becomes:
cos ^2 x+2sinxcosx = 0
cosx(cosx+sinx) =0
Or cosx = 0 or Osx+sinx = 0 Or cosx+sinx/cosx = 1+tanx = 0.
cos x = 0 gives x = pi/2 or x = -pi/2.
1+tanx = 0 gives: tanx = -1 Or s = 3pi/4 Or tanx = -pi/4.
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 = 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}
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