Determine x in interval[0,2pi]. cos5x+cos6x+cos7x=0

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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In order to determine x values, over 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 5x + cos 7x = 2 cos[(5x+7x)/2]cos[(5x-7x)/2]

cos 5x + cos 7x = 2cos(6x)cos(-2x)

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

So, instead of cos 5x + cos 7x, we'll substitute the sum by it's product:

2cos(6x)cos(2x)+ cos (6x) = 0

We'll factorize:

cos (6x)[2cos(2x)+1]=0

We'll put each factor as 0:

cos (6x)=0, is an elementary equation where


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


When k=0, x=pi/12

When k=1, x=5pi/12<2pi








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.

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neela | High School Teacher | (Level 3) Valedictorian

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To determine  cosx +cos6x+cos7x = 0 in (0,2pi)

Rearrange as:

(cos5x+cos7x)+cos6x = 0

2cos(5x+7x)/2 * cos (7x-5x)/2 +cos 6x = 0

2cos6x *cos x +cos6x = 0. Factorise cos6x.

2cos6x{cosx +1} =  0

cos6x = 0 , cosx +1 = 0

6x= 2kpi+or-pi/2  or x = pi

x = (2kp+pi/2)/6  = (4pi+1)pi/12  or x = (4k-1)pi/1`2 for k =0,1,2,...5

x = pi.