Question

Решите уравнение: arccos^2x-8arccosx+15=0

Answer 1

Arccosx=t
t^2-8t+15=0 t=3 t=5 но ни одно из них не подходит поскольку область значений косинуса
[-1;1]

Answer 2

To solve the equation arccos^2x-8arccosx+15=0, we can use the following steps:

- Let y = arccos x, then the equation becomes y^2 - 8y + 15 = 0. - Factor the quadratic equation as (y - 3)(y - 5) = 0. - Solve for y by setting each factor to zero: y - 3 = 0 or y - 5 = 0. - Get y = 3 or y = 5. - Substitute back y = arccos x and use the inverse property of arccos to get x = cos y. - Get x = cos 3 or x = cos 5. - Use a calculator or a table of arccos values to find the approximate values of x. - Get x ≈ 0.99 or x ≈ 0.28.

Therefore, the solutions of the equation are x ≈ 0.99 or x ≈ 0.28. You can check your answer by plugging these values into the original equation and see if they satisfy it.

For more information about arccos and its properties, you can refer to the web search results I found for you . You can also use an online equation solver to verify your answer. I hope this helps.