Question

Cos2x + sin²x = 0,5 [-7π/2; -2π]

Answer 1

Cos2x=cos^x-sin^x=1-2sin^x. Имеем: 1-2sin^x + sin^x=0,5. sin^x=1/2. Значит sin x = корень из 2 поделить на 2 и sin x = - корень из 2 поделить на 2. Тогда x = п/4 + 2пk, x=3п/4 +2пk, x= - п/4 + 2пk, x= - 3п/4 + 2пk. На заданный промежуток выпадают только - 13п/4, - 11п/4, -- 9п/4.

Answer 2

To solve the equation cos(2x) + sin^2(x) = 0.5 in the interval [-7π/2, -2π], you can follow these steps:

1. Rewrite the equation using trigonometric identities: cos(2x) + (1 - cos^2(x)) = 0.5

2. Substitute u = cos(x): cos(2x) + (1 - u^2) = 0.5

3. Solve for cos(2x): cos(2x) = 0.5 - (1 - u^2)

4. Simplify: cos(2x) = u^2 - 0.5

5. Now, we need to find the values of u in the given interval. In the interval [-7π/2, -2π], u (cos(x)) will vary from -1 to -1/2.

6. Solve for u^2 - 0.5 in the range [-1, -1/2]: -1 ≤ u^2 - 0.5 ≤ -0.5

7. We need to find the values of u that satisfy this inequality. Let's start by solving for u^2 - 0.5 ≥ -0.5: u^2 - 0.5 ≥ -0.5 u^2 ≥ 0

8. Since u^2 is always greater than or equal to zero, any value of u in the range [-1, -1/2] will satisfy this inequality.

9. So, all values of u in the interval [-1, -1/2] will work.

10. Now, we need to find the corresponding values of x. Since u = cos(x), we have: -1 ≤ cos(x) ≤ -1/2

11. To find the values of x, you can take the inverse cosine (arccos) of these values while considering the correct quadrant (quadrant II where cosine is negative):

    -1 ≤ cos(x) ≤ -1/2 π ≤ x ≤ 2π/3

12. However, we need to ensure that these values also fall within the given interval [-7π/2, -2π]. The interval [-7π/2, -2π] is equivalent to [-7π, -4π], so we have:

    π ≤ x ≤ 2π/3

13. The values of x that satisfy the equation cos(2x) + sin^2(x) = 0.5 in the interval [-7π/2, -2π] are: π ≤ x ≤ 2π/3

These are the solutions for the given equation in the specified interval.