cos3x+cos7x=cos2x what is x?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should convert the sum `cos3x+cos7x`  into a product such that:

`cos3x+cos7x = 2cos((3x+7x)/2)cos((3x-7x)/2)`

`cos3x+cos7x = 2cos 5x*cos (-2x)`

You should remember that the cosine function is even, hence,`cos(-alpha) = cos alpha` .

`cos3x+cos7x = 2cos 5x*cos 2x`

Substituting `2cos 5x*cos 2x`  for `cos3x+cos7x`  yields:

`2cos 5x*cos 2x = cos 2x`

You need to move all terms to the left such that:

`2cos 5x*cos 2x- cos 2x = 0`

You need to factor out `cos 2x`  such that:

`cos 2x*(2cos 5x - 1) = 0`

You need to solve the equations `cos 2x = 0`  and `2cos 5x - 1= 0`  such that:

`cos 2x = 0 =gt 2x = +-cos^(-1)(0) + 2npi`

`2x = 2npi =gt x = npi`

`2cos 5x - 1 = 0 =gt 2cos 5x= 1 =gt cos 5x = 1/2`

`5x = +-cos^(-1)(1/2)+ 2npi`

`5x = +-pi/3 + 2npi =gt x = +-pi/15 +(2npi)/15`

Hence, evaluating the general solutions to the given equation yields `x = npi`  and `x = +-pi/15 + (2npi)/5` .

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jeew-m | College Teacher | (Level 1) Educator Emeritus

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cos3x+cos7x=cos2x

cos3x+cos7x-cos2x = 0

 We know that `cos A +cos B = 2cos((A+B)/2)*cos ((A-B)/2)`

 

cos7x+cos3x-cos2x = 0

2cos((7x+3x)/2)*cos((7x-3x)/2) - cos2x = 0

2cos5x * cos2x-cos2x = 0

cos2x(2cos5x-1) = 0

 

The solution of this expression is cos2x = 0 and cos5x = 1/2

 

cos2x = 0

cos2x = `cos(pi/2)`

General solution for cosines is `x = 2n*pi+-alpha`

cos2x = `cos (pi/2)`

     2x = `2n*pi+-pi/2`    where  `n in Z`

       x = `n*pi+-pi/4`

 

When n=-1; x = `-3pi/4` or `-5pi/4`

When n= 0; x =`pi/4` or `-pi/4`

When n=-1; x =`5pi/4` or `3pi/4`

 

cos5x = 1/2

cos5x = `cos(pi/3)`

     5x = `2m*pi+-pi/3`    where  `m in Z`

       x = `2m*pi/5+-pi/15`

 

When m=-1; x = `-1/3*pi` or `-7/15*pi`

When m= 0; x = `pi/15` or `-pi/15`

When m=-1; x = `7/15*pi` or `1/3*pi`

 

 

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bhm001 | Student, Undergraduate | eNotes Newbie

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we know

Cos(A+B)=CosA.CosB-SinA.SinB 

Cos(A-B)=CosA.CosB+SinA.SinB

Cos3x=Cos(5x-2x)
Cos7x=Cos(5x+2x)
So:
(Cos5x.Cos2x+Sin5x.Sin2x)+(Cos5x.Cos2x-Sin5x.Sin2x) =Cos2X
2*Cos5x.Cos2x=Cos2x

1)Cos2x=0 

 

2)2Cos5x=1

cos5x = 1/2

(Answers are in the above post BTY the first answer is not complete the nP should be nP/2)

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