# What is x if (1+cos2x)/2=sin(-2x)?

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### 2 Answers

We have to solve (1 + cos 2x) / 2 = sin (-2x)

(1 + cos 2x) / 2 = sin (-2x)

=> -(1 + cos 2x) / 2 = sin (2x)

=> -( 1 + (cos x)^2 - (sin x)^2)/ 2 = 2 sin x cos x

=> -1 - (cos x)^2 + (sin x)^2 = 4 sin x cos x

=> -(cos x)^2 - (sin x)^2 - (cos x)^2 + (sin x)^2 =4 sin x cos x

=> - 2(cos x)^2 = 4 sin x cos x

=> -2 cos x = 4 sin x

=> sin x / cos x = -2/4

=> tan x = -1/2

x = arc tan (-1/2)

**Therefore x = arc tan ( -1/2) + n*pi**

We'll recognize in the first term the formula of the half angle.

(1+cos2x)/2 = (cos x)^2

We also know that the sine function is odd, so sin(-2x) = - sin 2x.

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

We'll add 2sin x * cos x both sides:

(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}**