Question

| sin x + cos x |< √ 2/2

Answer 1

Ответ на картинке внизу страницы

Answer 2

Let's start by looking at the different cases when the inequality could hold:

Case 1: sin x + cos x > 0 If sin x + cos x > 0, then |sin x + cos x| = sin x + cos x, and we can square both sides of the inequality to get:

(sin x + cos x)^2 < 2/2

Simplifying this, we get:

sin^2 x + 2 sin x cos x + cos^2 x < 1

Using the identity sin^2 x + cos^2 x = 1, we can simplify further:

2 sin x cos x < 0

This inequality holds when sin x and cos x have opposite signs, which means that x is in the second or fourth quadrant. Therefore, the inequality holds for x in the interval [3π/4, 5π/4].

Case 2: sin x + cos x < 0 If sin x + cos x < 0, then |sin x + cos x| = -(sin x + cos x), and we can square both sides of the inequality to get:

(sin x + cos x)^2 < 2/2

Simplifying this, we get:

sin^2 x + 2 sin x cos x + cos^2 x < 1

Using the identity sin^2 x + cos^2 x = 1, we can simplify further:

2 sin x cos x > 0

This inequality holds when sin x and cos x have the same sign, which means that x is in the first or third quadrant. Therefore, the inequality holds for x in the interval [π/4, 3π/4] or [5π/4, 7π/4].

Combining both cases, we get that the inequality holds for x in the intervals [3π/4, 5π/4] or [π/4, 3π/4] or [5π/4, 7π/4].