Question

-2sin2x < v3 V-корень

Answer 1

Ответ:

2sin2x=&2

sin2x=&2/2

2x=п/4+2пk x=п/8+пk, k₸z

2x=3а/4+2пk x=3п/8+пk, k₸z

Answer 2

It seems like you have an inequality involving trigonometric functions and a square root. Let me try to help you solve it step by step.

The inequality is: -2sin(2x) < √3√(-корень)

I'm not sure what you mean by "√3√(-корень)" as it's not a standard mathematical notation. However, I'll assume you mean "√3" for simplicity.

So, the modified inequality is: -2sin(2x) < √3

Now, let's solve for x:

1. First, divide both sides by -2. Remember that when you divide an inequality by a negative number, you need to reverse the inequality sign: sin(2x) > -√3/2

2. Now, consider the range of the sine function. The sine function has values between -1 and 1, inclusive. Therefore, -1 ≤ sin(2x) ≤ 1.

3. Since sin(2x) is always between -1 and 1, we can conclude that: -1 ≤ sin(2x) < -√3/2

4. Now, find the values of x that satisfy this inequality. To do this, we need to consider where sin(2x) is less than -√3/2.

5. The sine function is negative in the third and fourth quadrants of the unit circle. In those quadrants, sin(2x) is negative.

6. Additionally, sin(60 degrees) = √3/2, and since we want sin(2x) to be less than -√3/2, we need 2x to be in the third or fourth quadrant.

7. In the third quadrant, 2x lies between 180 and 270 degrees (π and 3π/2 radians), and in the fourth quadrant, it lies between 270 and 360 degrees (3π/2 and 2π radians).

So, the solution to the inequality is: 180 degrees ≤ 2x < 360 degrees π radians ≤ 2x < 2π radians

Now, if you want to find the solution for x in the range [0, 2π], you can divide both sides of the inequality by 2:

90 degrees ≤ x < 180 degrees π/2 radians ≤ x < π radians

So, the solution to the inequality in the range [0, 2π] is π/2 ≤ x < π.