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Solve the equation on the interval [0,2π) .

4cos^2x−2cosx=0

 Apr 10, 2020
 #1
avatar+485 
+2

Name cosx = y

 

4y2 - 2y = 0

Dividing by 2 on both sides, we get:

2y2 - y = 0

Dividing by y on both sides, we get:

2y -1 = 0

Adding 1 on both sides, weget:

2y = 1

y = 1/2

Substituting back in:

cosx = 1/2

the values for which this is true in the interval [0,2pi) are 

pi/6 and 5pi/6

 Apr 10, 2020
 #2
avatar+111329 
+3

4cos^2(x)  - 2cos (x)  =  0

 

2cos (x)  [  2cos(x) - 1 ]  =  0

 

So either

 

2 cos x  = 0

cos x  = 0

And  this happens at  x  = pi/2

 

Also

 

2cos x - 1  = 0

cos x  = 1/2

And this happens  at  x =  pi / 3  and  5pi/3 

 

cool cool cool

 Apr 10, 2020
 #3
avatar+4569 
+1

It's better to remember the unit circle for this one...

 

pi / 3 + 2pi (n) 

 

5pi/3  + 2pi(n)

 Apr 10, 2020

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