If cos x + cos (3x) + cos (5x) = 0, what can x be between 0 and 2pi?

In order to see if x can be situated in the interval [0,2pi], we'll try to solve the expression above, in order to find out the value of the unknown x.

For solving the expression, it will be useful to group the firs and the last terms together, so that the sum can be transformed into a product, after the following formula:

cos x + cos y= 2cos[x+y)/2]cos [(x-y)/2]

cos x + cos (5x)= 2 cos[(x+5x)/2]cos[(x-5x)/2]

cos x + cos (5x)= 2cos(3x)cos(2x)

So, instead of cos x + cos (5x), we'll substitute the sum with it's product:

2cos(3x)cos(2x)+ cos (3x)=0

We've noted that we have cos (3x) as common factor, so the expression could be written in this way:

cos (3x)[2cos(2x)+1]=0

As we can see, the expression above it is a product between 2 factors and it is equal to 0. That means:

cos (3x)=0, elementary equation where

3x=arccos(0)+2kpi

3x=pi/2 + 2kpi

x=pi/6+2kpi/3

When k=0, x=pi/6

When k=1, x=2pi/3 -(pi/6)=3pi/6=pi/2<2pi

[2cos(2x)+1]=0

2cos(2x)=-1

cos(2x)=(-1/2)

2x=arccos(-1/2)+2kpi

2x=pi-arccos(1/2)+2kpi

2x=pi-(pi/3)+2kpi

x=(pi/2)-(pi/6)+kpi

x=(3pi-pi)/6 +(kpi)

x=2pi/6+ (kpi)

x=pi/3 +kpi

When k=0, x=pi/3<2pi 

When k=1, x=pi/3+pi=4pi/3<2pi(240 degrees are found in the third quadrant)

When k=2, x=-pi/3+2pi=5pi/3<2pi(300 degrees are found in the fourth quadrant)