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

3 Answers | Add Yours

hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

(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)

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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}

We’ve answered 318,991 questions. We can answer yours, too.

Ask a question