Question

Sin2x - 3sinxcosx +cos2x= -1 PS: в sin2x двойка 一 степень

Answer 1

Ответ:

решений нет

Объяснение:

решение во вложении

Answer 2

To solve the equation sin^2(x) - 3sin(x)cos(x) + cos^2(x) = -1, we can use the trigonometric identity:

sin^2(x) + cos^2(x) = 1.

Using this identity, we can rewrite the equation as:

1 - 3sin(x)cos(x) = -1.

Now, let's isolate the term with sin(x)cos(x):

-3sin(x)cos(x) = -2.

Next, divide both sides by -3:

sin(x)cos(x) = 2/3.

Now, we can use the double angle identity for sine:

sin(2x) = 2sin(x)cos(x).

Substitute sin(x)cos(x) with 2/3:

sin(2x) = 2 * 2/3,

sin(2x) = 4/3.

However, there is a problem with this result. Sine values always lie in the range [-1, 1], so sin(2x) cannot be equal to 4/3. This means there is no solution to the given equation. It's possible there might have been a mistake in the equation or problem setup. Double-check the original equation or let me know if you need further assistance!