SOLUTION: what angles in between pi and 2pi satisfy the equation cos(4x)=0

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Question 803454: what angles in between pi and 2pi satisfy the equation cos(4x)=0
Answer by Edwin McCravy(20081) About Me  (Show Source):
You can put this solution on YOUR website!
cos(4x)=0

The angles with cosine 0 are all the odd multiples of pi%2F2

4x = %282n%2B1%29pi%2F2

 x = %282n%2B1%29pi%2F2÷4
 x = %282n%2B1%29pi%2F2%22%22%2A%22%221%2F4
 x = %282n%2B1%29pi%2F8 

 pi%3C=%282n%2B1%29pi%2F8%3C=2pi

Divide all three sides by p

 1%3C=%282n%2B1%29%2F8%3C=2

Multiply all three sides by 8 to clear of fractions:

 8%3C=2n%2B1%3C=16

Subtract 1 from all three sides

 7%3C=2n%3C=15

Divide all three sides by 2

 3.5%3C=n%3C=7.5

Since n is an integer
 
  4%3C=n%3C=7, 

  n ∈ {4,5,6,7}

Answers:  x = %282%284%29%2B1%29pi%2F8 = %288%2B1%29pi%2F8 = 9pi%2F8 
          x = %282%285%29%2B1%29pi%2F8 = %2810%2B1%29pi%2F8 = 11pi%2F8 
          x = %282%286%29%2B1%29pi%2F8 = %2812%2B1%29pi%2F8 = 13pi%2F8 
          x = %282%287%29%2B1%29pi%2F8 = %2814%2B1%29pi%2F8 = 15pi%2F8 

Edwin