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