Tính giá trị biêu thức sau

To solve the equation 5cos(4x) - 3cos(2x) + 4 = 0, we can use the double angle formula for cosine:

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

Substituting this into the equation, we get:

5cos(4x) - 3(2cos^2(x) - 1) + 4 = 0

Simplifying further:

5cos(4x) - 6cos^2(x) + 3 + 4 = 0

Rearranging the terms:

5cos(4x) - 6cos^2(x) + 7 = 0

Now, let's solve for cos(4x) using a substitution:

Let y = cos(x)

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

Substituting this back into the equation:

5(8y^4 - 8y^2 + 1) - 6y^2 + 7 = 0

Simplifying further:

40y^4 - 38y^2 + 12 = 0

Now, let's solve this quadratic equation for y^2:

Using the quadratic formula:

y^2 = (-(-38) ± √((-38)^2 - 4(40)(12))) / (2(40))
= (38 ± √(1444 - 1920)) / 80
= (38 ± √(-476)) / 80

Since the discriminant is negative, there are no real solutions for y^2. Therefore, there are no real solutions for cos(x) and consequently no real solutions for x.