# Given o <= x <= 2pi solve equation cos4x=1.

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### 2 Answers

We have to solve cos 4x = 1 for values of x that satisfy 0<=x<=2*pi

cos 4x = cos 2*2x

=> 1 - 2(sin 2x)^2

=> 1 - 2*(2*cos x * sin x)^2

=> 1 - 8*(cos x)^2*(sin x)^2

=> 1 - 8*(1 - (sin x)^2)(sin x)^2

let (sin x)^2 = y

=> 1 - 8*(1 - y)y = 1

=> (1 - y)y = 0

=> y = 0 and y = 1

sin x = 0

=> x = arc sin 0

=> x = 0, pi , 2*pi

(sin x)^2 = 1

=> x = arc sin 1 and x = arc sin (-1)

=> x = pi/2 and x = 3*pi/2

**The required solution of the equation is x = {0 , pi/2, pi , 3*pi/2, 2*pi}**

Since x belongs to the closed interval [0 ; 2pi], then 4*x belongs to the interval [0 ; 8pi].

cos 4x = 1 if and only if 4x has the following values from the interval

[0 ; 8pi]: {0 ; 2pi ; 4pi ; 6pi ; 8pi}.

Now, we'll conclude that x has the followings values: {0/4 ; 2pi/4 ; 4pi/4 ; 6pi/4 ; 8pi/4} = {0 ; pi/2 ; pi ; 3pi/2 ; 2pi}.

Another method to solve the problem is to put 1 = cos 0.

**The values of x, in the range [0 ; 2pi], are: {0 ; pi/2 ; pi ; 3pi/2 ; 2pi}.**