Question

-2sin(n/9-x/9)= 2sin(n/12-x/12)=-1 2sin^2-3sinx-2=0

Answer 1

Решение смотри на фотографии

Answer 2

Let's take each equation separately and solve them:

Equation 1:

-2sin(n/9-x/9) = 2sin(n/12-x/12)

Divide both sides by 2:

-sin(n/9-x/9) = sin(n/12-x/12)

Using the identity sin(a-b) = sin(a)cos(b) - cos(a)sin(b), we can rewrite the equation as:

sin(n/12)cos(x/12) - cos(n/12)sin(x/12) = -sin(n/9)cos(x/9) + cos(n/9)sin(x/9)

Rearranging and simplifying:

(sin(n/12)cos(x/12) + sin(n/9)cos(x/9)) = (cos(n/12)sin(x/12) + cos(n/9)sin(x/9))

Using the identity sin(a+b) = sin(a)cos(b) + cos(a)sin(b), we can further simplify the equation as:

sin((n/12)+(n/9))cos(x/12) = sin((n/12)+(n/9))sin(x/12)

Since sin((n/12)+(n/9)) is not equal to zero, we can divide both sides by it:

cos(x/12) = sin(x/12)

This can be rewritten as:

tan(x/12) = 1

Solving for x, we get:

x = 12tan⁻¹(1) + 24kπ, where k is an integer.

Therefore, the solutions for the first equation are:

x = 12(π/4) + 24kπ or x = 12(5π/4) + 24kπ, where k is an integer.

Equation 2:

2sin^2x - 3sinx - 2 = 0

This is a quadratic equation in sinx. We can solve it using the quadratic formula:

sinx = (3 ± √17)/4

Therefore, the solutions for the second equation are:

x = sin⁻¹[(3 + √17)/4] + 2kπ or x = sin⁻¹[(3 - √17)/4] + 2kπ, where k is an integer.