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

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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)

2cosx*cos(3x)+1 = 0

2cosx[(4cosx)^3-3cosx)+1 = 0 , as cos3x =cos(2x+x)=2cos2xcosx-sin2xsinx=(2(2cosx)^2-1)*(cosx)-2sin2xsinx=2(cosx)^3-cosx-2(sinx)^2cosx=2(cosx)^3-2[1-(cosx)^2]cosx) = 4( cosx)^3-3cosx.

8(cosx)^4-6cos^2x+1 = 0 is a quadratic in (cosx)^2. We factorise the left expression which is equal to zero.

[4(cosx)^2-1)][(2cosx)^2-1]=0

So, 4(cosx)^2 = 1 or 2(cosx)^2 = 1

cosx = sqrt(1/4) = +(1/2) or -(1/2) or

cosx = +sqrt(1/2) or -sqrt(1/2)

From cosx =1/2 , x = p/3 or x = -p/3 = 5p/3 , where p is pi .

From cosx= -1/2 , p = 2p/3 or -2p/3 = 4p/3,

From cosx = sqrt(1/2) , x = p/4 or -p/4 = 7p/4 and

From cosx = -sqrt(1/2) , x=3p/4 or -3p/4 = 5p/4 all being in radian measures.

Therefore,

x=p/3 or 5p/3 or 60 degree or 300 degree

x=2p/3 or 4p/3 or 12 degree or 240 degree

x=p/4 or 7p/4 or 45 degree or 315 degree.