Question

1) sin квадрат 2x ≥ 3sin2x; 2)2sin квадрат x-cos квадрат x-sinx cosx > 0

Answer 1

Решение смотри во вложении.

Answer 2

1. To solve the inequality sin²2x ≥ 3sin2x, we can first factor out sin2x from both sides:

sin2x(sin2x - 3) ≥ 0

Now we can use the zero product property and solve for sin2x - 3 = 0 and sin2x ≥ 0 separately:

sin2x - 3 = 0 sin2x = 3

This equation has no real solutions since the sine function's range is between -1 and 1.

sin2x ≥ 0

This inequality is true for all values of x, so the solution to the original inequality is:

sin²2x ≥ 3sin2x is true for all x.

2. To solve the inequality 2sin²x - cos²x - sinx cosx > 0, we can first use the identity cos²x = 1 - sin²x to simplify:

2sin²x - (1 - sin²x) - sinx cosx > 0

Rearranging, we get:

3sin²x + sinx cosx - 1 > 0

Now we can use the quadratic formula to solve for sinx:

sinx = (-b ± sqrt(b² - 4ac)) / 2a

where a = 3, b = 1, and c = -1. Plugging these values in, we get:

sinx = (-1 ± sqrt(13)) / 6

Since sinx cannot be greater than 1 or less than -1, we can ignore the solution sinx = (-1 - sqrt(13)) / 6. Thus, the solution to the inequality is:

sinx > (-1 + sqrt(13)) / 6.