Question

6cos^2x-13sin x-13=0 тригонометрические уравнения

Answer 1

- 1 1/6 не подходит под ОДЗ

Answer 2

To solve the trigonometric equation 6cos^2x-13sin x-13=0, we can use the identity cos^2(x) + sin^2(x) = 1 to express cos^2(x) in terms of sin(x):

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

Substituting this into the original equation, we get:

6(1-sin^2(x)) - 13sin(x) - 13 = 0

Expanding and rearranging, we get:

-6sin^2(x) - 13sin(x) + 1 = 0

We can solve for sin(x) using the quadratic formula:

sin(x) = [-(-13) ± √((-13)^2 - 4(-6)(1))] / (2(-6)) sin(x) = [13 ± √(265)] / 12

Now, we can find the corresponding values of cos(x) using the identity cos^2(x) + sin^2(x) = 1:

cos(x) = ±√[1 - sin^2(x)]

Plugging in the values of sin(x), we get:

cos(x) = ±√[1 - (13 ± √(265))^2 / 144]

Therefore, the solutions to the trigonometric equation are:

sin(x) = (13 + √265) / 12, cos(x) = ±√[1 - (13 + √265)^2 / 144]

sin(x) = (13 - √265) / 12, cos(x) = ±√[1 - (13 - √265)^2 / 144]