Question

Sin2x=-1объяснитеподробнее​

Answer 1

$sin2x=-1$

Синус какого-то числа равен -1 в нижней точке на единичной окружности

Эта точка = $\dfrac{3\pi }{2}$ , но синус принимает это значение каждый оборот круга, то есть каждые 2π радиан

Значит все точки, у которых синус -1 это  $\dfrac{3\pi }{2}+2\pi n$ , где n целое число

Тогда $2x=\frac{3\pi }{2}+2\pi n\\ \\ x=\frac{3\pi }{4} +\pi n$ , n всё ещё целое число

Answer 2

To solve the equation sin(2x) = -1, we need to find the value(s) of x that satisfy this equation. In other words, we want to find the angle (or angles) whose sine of twice the angle equals -1.

To do this, we can use the properties of the sine function and some trigonometric identities.

Recall that the sine function is periodic, with a period of 2π. This means that for any angle θ, sin(θ) = sin(θ + 2π). Additionally, we know that sin(π) = 0 and sin(3π/2) = -1.

Now, let's focus on the equation sin(2x) = -1. We need to find angles whose double has a sine value of -1.

Step 1: Write the equation in terms of a double angle formula. The double angle formula for sine is: sin(2θ) = 2sin(θ)cos(θ).

Using this formula, we can rewrite the equation as follows: 2sin(x)cos(x) = -1.

Step 2: Find the values of sin(x) and cos(x) that satisfy the equation. Since sin(π) = 0 and sin(3π/2) = -1, we know that sin(x) can be -1 for some value of x.

Now, let's find the corresponding value of cos(x). We can use the Pythagorean identity for the sine and cosine functions, which states: sin^2(θ) + cos^2(θ) = 1.

Since sin(x) = -1, we have: (-1)^2 + cos^2(x) = 1, 1 + cos^2(x) = 1, cos^2(x) = 0.

Taking the square root of both sides, we get: cos(x) = ±√0, cos(x) = 0.

Step 3: Determine the possible values of x. We have found that sin(x) can be -1, and cos(x) can be 0 to satisfy the equation sin(2x) = -1.

Now, let's find the corresponding values of x:

For sin(x) = -1 and cos(x) = 0: x = π/2 + kπ, where k is an integer.

For sin(x) = -1 and cos(x) = 0 (using the negative square root): x = 3π/2 + kπ, where k is an integer.

So, the solutions to the equation sin(2x) = -1 are given by: x = π/2 + kπ, and x = 3π/2 + kπ, where k is an integer.

These are the values of x for which sin(2x) equals -1. There are infinitely many solutions because sine is a periodic function, and for any integer value of k, the angle (π/2 + kπ) and (3π/2 + kπ) will have the same sine value when doubled.