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

neela | Student

cosx+cos3x+cos5x  = 0 is an equation, the expresion on the left could be simplified and factorised into simple factors and each factor could then be equated to zero by 0 product rule and thus solved for x.



cosx+2cos[(3x+5x)/2]cos[(5x-3x)/2] = 0

cosx+2cos4x*cosx = 0

cosx{1+2cos4x} = 0

cosx=0  or 1+2cos4x = 0

Cosx =0, for x=90 and 270 degrees

1+2cos4x = 0 gives:

cos4x = -1/2.

4x = 120  or 4x = 240 .  General solution  4x= (n*360 +or- 120) degree. So,

x = n*90+30 or x=n*90-30 degree.

Where n is an integer.For n=0,1,2,3 you get x values in degrees .

The angles,30 or 120  or 150  or 210 ,240, 300 and 330 degrees  hold good  in the interval 0 to 360 dgrees.(p/6,4p/6, 5p/3,7p/6,8p/6,10p/6 and 11p/6 radians)

giorgiana1976 | Student

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=pi/2 + 2kpi


When k=0, x=pi/6

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








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)

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