Question

I know the answer to this problem but I have no idea how to get there.


Solve for all values of x if 0 is less than or equal to x, which is less than or equal to 2pie.
(cosine squared of 4x) + (cosine of 4x)=0

The answer is:
x= pie/8, 3pie/8, 5pie/8, 7pie/8, 9pie/8, 11pie/8, 13pie/8, 15pie/8, pie/4, 3pie/4, 5pie/4, 7pie/4

Answer 1

it factors to
cos4x(cos4x + 1) = 0
cos4x = 0 or cos4x = -1
4x = pi/2 or 4x = 3pi/2 or 4x = pi
look at the cosine curve for these

then x = pi/8 or x = 3pi/8 or x = pi/4

we know the period of cos4x is 2pi/4 or pi/2
so by adding pi/2 to any of the above 3 answers, we obtain a new answer.
Do this until you pass the domain of 2pi.

e.g. 3pi/8 + pi/2
= 3pi/8 + 4pi/8
= 7pi/8 , one of the answers

then 7pi/8 + pi/2
= 7pi/8 + 4pi/8
= 11pi/8 , another answer.

I am sure you get the idea.