Question

Cos(x/2) + sin(x/2) = -1 3sin(x) - 4cos(x) = 5

Answer 1

Ответ:

sinx-2sinx-1+корень 3 sinx+sin4П/3

Объяснение:

Answer 2

To solve the given system of equations:

Equation 1: cos(x/2) + sin(x/2) = -1 Equation 2: 3sin(x) - 4cos(x) = 5

Let's solve these equations step by step.

First, we'll square both sides of Equation 1 to eliminate the square root:

(cos(x/2))^2 + 2sin(x/2)cos(x/2) + (sin(x/2))^2 = 1

Simplifying, we have:

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

Now, let's focus on Equation 2:

3sin(x) - 4cos(x) = 5

Dividing the equation by 5 to simplify, we get:

(3/5)sin(x) - (4/5)cos(x) = 1

Now, let's express sin(x) and cos(x) in terms of a single trigonometric function to simplify further. We can use the identity: sin^2(x) + cos^2(x) = 1.

Dividing Equation 2 by sqrt((3/5)^2 + (-4/5)^2) = sqrt(9/25 + 16/25) = sqrt(25/25) = 1, we have:

(3/5)sin(x) - (4/5)cos(x) = 1

Multiplying both sides by sqrt(25), we get:

3sin(x) - 4cos(x) = 5

Notice that this is the same equation as Equation 2. This implies that Equation 1 and Equation 2 are equivalent, and we can solve this new equation instead.

Now, we have:

3sin(x) - 4cos(x) = 5

Let's solve this equation using the substitution method:

From Equation 1 (sin(x)cos(x) = 0), we have two possibilities:

1. sin(x) = 0, which implies x = 0 or x = π (or any integer multiple of π).

2. cos(x) = 0, which implies x = π/2 or x = 3π/2 (or any odd multiple of π/2).

Now, substitute these values into Equation 2:

For x = 0: 3sin(0) - 4cos(0) = 5 0 - 4(1) = 5 -4 = 5 (Not true)

For x = π: 3sin(π) - 4cos(π) = 5 0 - 4(-1) = 5 4 = 5 (Not true)

For x = π/2: 3sin(π/2) - 4cos(π/2) = 5 3(1) - 4(0) = 5 3 = 5 (Not true)

For x = 3π/2: 3sin(3π/2) - 4cos(3π/2) = 5 3(-1) - 4(0) = 5 -3 = 5 (Not true)

Therefore, there are no solutions to the given system of equations.

If you have any further questions, please, let me know!