Question

2(cos 4х)^− 6(cos 2х)^+ 1 = 0

Answer 1

Ответ:

x = ± π/6 + πm/2, m∈Z

Пошаговое объяснение:

Объяснение:

2cos²4x - 6cos²2x + 1 = 0

2(2cos²2x - 1)² - 6cos²2x + 1 = 0

2(4cos⁴2x - 4cos²2x + 1) - 6cos²2x + 1 = 0

8cos⁴2x - 8cos²2x + 2 - 6cos²2x + 1 = 0

8cos⁴2x - 14cos²2x + 3 = 0

cos²2x = t

8t² - 14t + 3 = 0

D/4 = 7² - 24 = 49 - 24 = 25

t_1=\dfrac{7+5}{8}=\dfrac{3}{2}t

1

=

8

7+5

=

2

3

t_2=\dfrac{7-5}{8}=\dfrac{1}{4}t

2

=

8

7−5

=

4

1

cos²2x = 3/2 - нет корней

или

cos²2x = 1/4

cos2x = ± 1/2

cos2x = 1/2 cos2x = - 1/2

2x = ± π/3 + 2πn, n∈Z 2x = ± 2π/3 + 2πk, k∈Z

Корни можно объединить:

2x = ± π/3 + πm

x = ± π/6 + πm/2, m∈Z

Answer 2

To solve the equation 2(cos(4x))^(-6)(cos(2x))^2 + 1 = 0, we can first simplify it by applying some trigonometric identities. Let's start:

Let's denote y = cos(2x). The equation now becomes:

2(cos(4x))^(-6)y^2 + 1 = 0

To simplify further, we can use the double-angle formula for cosine:

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

Now let's substitute this back into the equation:

2(2cos^2(2x) - 1)^(-6)y^2 + 1 = 0

To proceed, let's focus on the first term (2cos^2(2x) - 1)^(-6):

Using the identity cos^2(2x) = (1 + cos(4x))/2, we can rewrite it as:

(2(1 + cos(4x))/2 - 1)^(-6)

Simplifying further:

(2 + 2cos(4x) - 1)^(-6) = (1 + 2cos(4x))^(-6)

Substituting this back into the equation, we have:

2(1 + 2cos(4x))^(-6)y^2 + 1 = 0

Now we have a quadratic equation in terms of y:

2(1 + 2cos(4x))^(-6)y^2 + 1 = 0

To solve this equation, we can isolate y^2:

2(1 + 2cos(4x))^(-6)y^2 = -1

y^2 = -1 / [2(1 + 2cos(4x))^(-6)]

Since y = cos(2x), we have:

cos^2(2x) = -1 / [2(1 + 2cos(4x))^(-6)]

Now we can solve for cos(2x):

cos(2x) = ±sqrt(-1 / [2(1 + 2cos(4x))^(-6)])

The equation does not have a straightforward solution due to the presence of a complex number (sqrt(-1)). Please double-check the equation or provide more information if there is a specific range or format for the solution you are looking for.