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

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.