# Which are x values in [0,2pi], for cos2x+cos3x+cos4x=0?

*print*Print*list*Cite

### 2 Answers

In order to find out which are x values, 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 first and the last terms together, so that the sum of the trigonometric functions to be transformed into a product, after the following formula:

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

cos 2x + cos 4x= 2 cos[(2x+4x)/2]cos[(2x-4x)/2]

cos 2x + cos 4x = 2cos(3x)cos(-2x)

But cos x is an even function, so cos(-2x)=cos 2x

So, instead of cos 2x + cos 4x, we'll substitute the sum with it's product:

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

We'll factorize:

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

We'll put each factor as 0:

cos (3x)=0, is an elementary equation where

3x=+/-arccos(0)+2*k*pi

3x=pi/2 + 2*k*pi

x=pi/6+2*k*pi/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)+2*k*pi

2x=pi-arccos(1/2)+2*k*pi

2x=pi-(pi/3)+2*k*pi

x=(pi/2)-(pi/6)+k*pi

x=(3pi-pi)/6 +k*pi

x=2pi/6+ k*pi

x=pi/3 +k*pi

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.

To find the solution for cos2x+cos3x+cos4x = 0 in the interval (0,2pi)

soltion:

We know that cosA +cosB = 2cos[(A+B)/2]*cos [(A-B)/2]. Using this to cos2x+cos4x , we get, 2cos[(2x+4x)/2] cos[(4x-co2x)/2] = 2cos3x*cosx. Replcing this in the given expression. we get:

2cos3x*cosx +cos3x = 0. Or

cos3x{2cosx+1} =0.

Therefore, cos3x = 0. Or 2cosx+1 = 0.

cos3x = 0, gives 3x = pi/2 or 3x = 3pi/2. x = Pi/6 Or x =(3pi/2)/3 = pi/2.

2cosx+1 = 0, gives: cosx = -1/2. Or x = 2pi/3 Or x = 4pi/3 .

So x = {Pi/6, pi/2, 2pi/3, 4pi/3 }