Question

2cosx-cos2x-cos^2 x=0

Answer 1

2*cosx-cos(2x)-cos²x=0
2*cosx-(cos²x-sin²x)-cos²x=0
2*cosx-cos²x+sin²x-cos²x=0
2*cosx-2*cos²x+1-cos²x=0
-3*cos²x+2*cosx+1=0  |×(-1)
3*cos²x-2*cosx-1=0
Пусть cosx=t   ⇒
3t²-2t-1=0      D=16      √D=4
t₁=1         ⇒     cosx=1             x₁=2πn
t₂=-1/3    ⇒     cosx=-1/3        x₂=arccos(-1/3)+2πn.

Answer 2

Cos2x = cos^2x - sin^2x = 2cos^2x - 1

2cosx - cos2x - cos^2x = 0
2cosx - 2cos^2x + 1 - cos^2x = 0
-3cos^2x + 2cosx + 1 = 0
Замена: cosx = t
-3t^2 + 2t + 1 = 0
3t^2 - 2t - 1 = 0
t = 1
t = -1/3

Обратная замена:
1) cosx = 1
x = 2пk, k принадлежит Z

2) cosx = -1/3
x = - arccos(-1/3) + 2пk
x = arccos(-1/3) + 2пk
k - целые числа

Ответ: -arccos(-1/3) + 2пk, arccos(-1/3) + 2пk, 2пk, k принадлежит Z

Answer 3

To solve the equation 2cos(x) - cos(2x) - cos^2(x) = 0, we can simplify it using trigonometric identities. Let's go step by step:

First, we'll rewrite cos(2x) using the double angle formula: cos(2x) = 2cos^2(x) - 1.

Now the equation becomes:

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

Next, distribute the negative sign:

2cos(x) - 2cos^2(x) + 1 - cos^2(x) = 0.

Combine like terms:

1 - cos^2(x) - 2cos^2(x) + 2cos(x) = 0.

Combine the similar terms with cos(x):

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

This is now a quadratic equation in terms of cos(x). Let's solve it by factoring or using the quadratic formula.

To make the equation easier to work with, let's substitute y = cos(x). The equation becomes:

-3y^2 + 2y + 1 = 0.

Now, we can factor this quadratic equation:

(-3y + 1)(y + 1) = 0.

Setting each factor equal to zero, we get:

-3y + 1 = 0 --> y = 1/3,

y + 1 = 0 --> y = -1.

Now substitute y back with cos(x):

cos(x) = 1/3 or cos(x) = -1.

To find the values of x, we can use the inverse cosine function (arccos) or use the unit circle.

1. For cos(x) = 1/3: Using the calculator or inverse cosine function, we find two solutions: x = arccos(1/3) + 2πn, x = -arccos(1/3) + 2πn,

where n is an integer representing additional solutions.

2. For cos(x) = -1: Using the calculator or inverse cosine function, we find one solution: x = π + 2πn,

where n is an integer representing additional solutions.

These are the general solutions to the equation 2cos(x) - cos(2x) - cos^2(x) = 0.