If 2*cosx*cos(3x)+1=0, what can x be between 0 and 2pi?

First of all, we'll write 2*cosx*cos(3x) as a sum, following the formula:

cos a+cos b= 2cos [(a+b)/2]cos[(a-b)/2]

So, 2*cosx*cos(3x)= cos a+ cos b

x=(a+b)/2 => a+b=2x

3x=(a-b)/2 => a-b=6x

a+b+a-b=2x+6x

2a=8x => a=4x

a+b-a+b=2x-6x

2b=-4x

b=-2x

Let's substitute now the found values for a and b:

2*cosx*cos(3x)= cos 4x+ cos 2x

The expression given is:

cos 4x+ cos 2x +1=0

Cos 4x= cos 2*(2x)=2[cos(2x)]^2-1

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

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

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

cos(2x)=0

2x=arccos 0 + 2kpi

2x=pi/2 + 2kpi

x=pi/4 + kpi

For k=0, x=pi/4=45 degrees(first quadrant)

For k=1, x=pi/4+pi=5pi/4=5*45=225 degrees(third quadrant)

But also, for k=1, x=pi-pi/4=3pi/4=3*45=135 (second quadrant)

2cos(2x)+1=0

2cos(2x)=-1

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

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

2x=pi/3+2kpi

x=pi/6+kpi

For k=0, x=pi/6=30degrees(first quadrant)

For k=1, x=pi/6+pi=7pi/6=7*30=210 degrees(third quadrant)

For k=1, x=pi-pi/6=5pi/6=5*30=150 degrees(second quadrant)