Question

3соs^2x-2sinxcosx-sin^2=0

Answer 1

Решение задания приложено

Answer 2

Решение представлено на фотографии. Ты еще забыла написать х в конце (sin ^2х).
Вроде бы решается так.

Answer 3

To solve the equation 3cos^2(x) - 2sin(x)cos(x) - sin^2(x) = 0, we can use trigonometric identities to simplify it and find the values of x.

Let's start by manipulating the equation:

3cos^2(x) - 2sin(x)cos(x) - sin^2(x) = 0

We know that cos^2(x) + sin^2(x) = 1, so we can substitute cos^2(x) with 1 - sin^2(x):

3(1 - sin^2(x)) - 2sin(x)cos(x) - sin^2(x) = 0

Expanding the equation:

3 - 3sin^2(x) - 2sin(x)cos(x) - sin^2(x) = 0

Combining like terms:

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

Rearranging the terms:

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

Now, let's factor out -2 from the first two terms:

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

Factoring out sin(x) from the parentheses:

-2sin(x)(2sin(x) + cos(x)) + 4 = 0

Dividing the equation by -2:

sin(x)(2sin(x) + cos(x)) - 2 = 0

So, we have:

sin(x)(2sin(x) + cos(x)) - 2 = 0

Now, we can solve this equation by setting each factor equal to zero:

sin(x) = 0 (Equation 1) 2sin(x) + cos(x) = 2 (Equation 2)

Let's solve Equation 1:

sin(x) = 0

The solutions for this equation are x = 0, π, 2π, 3π, etc.

Now, let's solve Equation 2:

2sin(x) + cos(x) = 2

To solve this equation, we can square both sides:

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

Expanding and simplifying:

4sin^2(x) + 4sin(x)cos(x) + cos^2(x) = 4

Using the trigonometric identity cos^2(x) + sin^2(x) = 1:

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

Simplifying further:

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

We can substitute 1 - cos^2(x) for sin^2(x):

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

Expanding:

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

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

Factor out cos(x):

cos(x)(-3cos(x) + 4sin(x)) = 0

So, we have two possibilities:

1. cos(x) = 0
2. -3cos(x) + 4sin(x) = 0

Let's solve each possibility:

1. cos(x) = 0

The solutions for this equation are x = π/2, 3π/2,