Question

Cos^2x-sin^2x-1-4cosx=0

Answer 1

Дано уравнение сos²x-sin²x-1-4cosx=0.
Заменим 1 = сos²x+sin²x.
сos²x-sin²x-сos²x-sin²x-4cosx=0.
-2sin²x-4cosx=0.
Заменим sin²x = 1 - сos²x.
-2 + 2сos²x-4cosx=0.  Сократим на 2.
сos²x - 2cosx - 1 = 0.   Заменим cosx = t.
Получаем квадратное уравнение t² - 2t - 1 = 0.
Квадратное уравнение, решаем относительно t: Ищем дискриминант:
D=(-2)^2-4*1*(-1)=4-4*(-1)=4-(-4)=4+4 = 8;Дискриминант больше 0, уравнение имеет 2 корня:
t_1=(√8-(-2))/(2*1)=(√8+2)/2=√8/2+2/2=√2+1 ≈ 2,41421 (это значение отбрасываем);t_2=(-√8-(-2))/(2*1)=(-√8+2)/2=-√8/2+2/2=-√+1 ≈ -0,41421.
Отсюда получаем ответ:
$x=arc cos(1- \sqrt{2} )+2 \pi k$, k ∈ Z.
$x=-arc cos(1- \sqrt{2})+2 \pi k,$ k ∈ Z.

Answer 2

Let's solve the given equation:

cos^2(x) - sin^2(x) - 1 - 4cos(x) = 0

First, we'll use the trigonometric identity:

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

Now, the equation becomes:

cos(2x) - 1 - 4cos(x) = 0

Next, let's rewrite the equation using only cosine terms:

2cos^2(x) - 1 - 4cos(x) = 0

Now, this is a quadratic equation in terms of cos(x). We can solve it by using substitution:

Let's make a substitution: u = cos(x)

Then the equation becomes:

2u^2 - 1 - 4u = 0

Now, we can solve this quadratic equation for u:

2u^2 - 4u - 1 = 0

To solve this quadratic equation, we can use the quadratic formula:

u = (-b ± √(b^2 - 4ac)) / 2a

where a = 2, b = -4, and c = -1.

u = [ -(-4) ± √((-4)^2 - 4 * 2 * (-1)) ] / (2 * 2)

u = (4 ± √(16 + 8)) / 4

u = (4 ± √24) / 4

Now, we have two possible solutions for u:

1. u = (4 + √24) / 4
2. u = (4 - √24) / 4

Now, we need to find the corresponding values of x:

1. u = (4 + √24) / 4 u = (4 + 2√6) / 4 u = 1 + √6 / 2

To find x, we use the inverse cosine (arccos) function:

x = arccos(u)

2. u = (4 - √24) / 4 u = (4 - 2√6) / 4 u = 1 - √6 / 2

x = arccos(u)

Now, the solutions for x are:

1. x = arccos(1 + √6 / 2)
2. x = arccos(1 - √6 / 2)

Keep in mind that these solutions may involve both the principal value and other periodic solutions since the cosine function has a periodic nature.