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`

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` .**

we know

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

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

(Cos5x.Cos2x+Sin5x.Sin2x)+(Cos5x.Cos2x-Sin5x.Sin2x) =Cos2X

2*Cos5x.Cos2x=Cos2x1)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)