(4-cos8x)(2+cos2x)=3How to solve this and what is the solution?

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

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We need to find x given (4 - cos 8x)(2 + cos 2x) = 3.

Now, cos 2x = 2(cos x)^2 - 1

cos 8x = 2(cos 4x)^2 - 1

=> 2(2(cos 2x)^2 - 1)^2 - 1

=> 2*[ 4 (cos 2x)^4 + 1 - 4(cos 2x)^2] - 1

=> 8 (cos 2x)^4 + 2 - 8(cos 2x)^2 - 1

Let cos 2x = y

=> 8y^4 - 8y^2 + 1

So (4 - cos 8x)(2 + cos 2x) = 3

=> (4 - (8 (cos 2x)^4 - 8(cos 2x)^2 + 1)))(2 + cos 2x) = 3

=> (4 - (8 y^4  - 8y^2 + 1))(2 + y) = 3

=> [ 4 - 8y^4 + 8y^2 - 1](2+y) = 3

=> (3 - 8y^4 + 8y^2)(2 + y) = 3

=> 6 - 16y^4 + 16y^2 + 3y - 8y^5 + 8y^3 - 3 =0

=>  16y^4 - 16y^2 - 3y + 8y^5 - 8y^3 - 3 =0

=> 8y^5 + 16y^4 - 8y^3 - 16y^2 -3y -3 =0

=> 8y^4( y+1) - 8y^2( y+1) - 3(y+1) = 0

=> (y+1)(8y^4 - 8y^2 - 3) =0

For y +1 = 0 , y = -1

For (8y^4 - 8y^2 - 3) =0

y1 = sqrt [8 + sqrt (64 + 96)] / 16

y1 = sqrt [(8 + sqrt 160) / 16]

y2 = - sqrt [(8 + sqrt 160) / 16]

y3 and y4 are complex.

The only valid root is y = -1, the other roots are complex or greater than 1 and can be ignored.

Therefore cos 2x = -1

We can find x by taking the arc cos of the values and dividing by 2.

x = 90 degrees.

Therefore x is equal to 90 degrees.

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

The problem has no solution for the following reasons:

Consider  the give equation (4-cos8x)(2+cos2x) = 15.

The LHS has the factors 4-cos8x and 2+cos2x.

The maximum value  we can assume  for the factor 4-cos8x is 5. and the maximum value we can assume for the factor 2+cos2x is 3.

Suppose the factor 4-cos8x has the maximum value of 5.

 Then  4-cos8x = 5 under the assumption.

Therefore cos8x = 4-5= -1 under the assumtion.

Therefore, 8x = (2n+or1)pi. So 2x = 2(2n+1)Pi/8 = (2n+1)pi/4  for n= 0,1, ...

Thererefore cos2x = cos(2n+1)pi/4 = p <1 for n = 0,1,2,.....

So the product (4-cos8x)(2+cos2x) < 5*(2+p)  <  5*3 = 15.

Therefore cos8x =  -1, is not possibe.

So for any other value of x , cos8x is less than > -1. and  4-cos8x is < 5. This implies the  second factor 2+cos2x must be more than 3. But 2+cosx > 3 is a contradiction as cosx > 1 is not possible.

Thus we proved that there is no solution to the  given problem.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll remove the brackets:

(4-cos8x)(2+cos2x)= 8 + 4cos 2x - 2cos 8x - cos 2x*cos 8x (1)

Cos 4x = cos 2*(2x) = (cos 2x)^2 - (sin 2x)^2

Cos 4x = 2(cos 2x)^2 - 1

cos 8x = cos 2*(4x) = (cos 4x)^2 - (sin 4x)^2

cos 8x = 2(cos 4x)^2 - 1

But Cos 4x = 2(cos 2x)^2 - 1:

cos 8x = 2[2(cos 2x)^2 - 1]^2 - 1 (2)

We'll substitute (2) in (1):

8 + 4cos 2x - 2cos 8x - cos 2x*cos 8x = 8 + 4cos 2x - 2{2[2(cos 2x)^2 - 1]^2 - 1} - cos 2x*[2[2(cos 2x)^2 - 1]^2 - 1]

8 + 4cos 2x - 2cos 8x - cos 2x*cos 8x = 8 + 4cos 2x - 2[4(cos 2x)^4 - 4(cos 2x)^2 + 1 - 1] - cos 2x*{2[4(cos 2x)^4 - 4(cos 2x)^2 + 1 - 1]}

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