Question

7cos^x-8sinxcosx+sin^x=0

Answer 1

Вот сверху своё вставь и всё

Answer 2

To solve the equation 7cos^x - 8sinx*cosx + sin^x = 0, we can simplify it and use trigonometric identities to rewrite the equation in terms of a single trigonometric function.

Let's start by using the Pythagorean identity sin^2(x) + cos^2(x) = 1 to rewrite the equation:

7cos^x - 8sinx*cosx + sin^x = 0

Dividing the entire equation by cos^x, we get:

7 - 8tanx + sin^x/cos^x = 0

Using the identity tanx = sinx/cosx, we can substitute it in the equation:

7 - 8tanx + tanx*sinx/cosx = 0

Now, let's multiply the entire equation by cosx to get rid of the denominators:

7cosx - 8sinx + tanx*sinx = 0

Next, we can rewrite tanxsinx as sin^2(x)/cosx using the identity tanxsinx = sin^2(x)/cosx:

7cosx - 8sinx + sin^2(x)/cosx = 0

Multiplying the entire equation by cosx, we obtain:

7cos^2(x) - 8sinx*cosx + sin^2(x) = 0

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

7 - 8sinx*cosx + 1 = 0

Rearranging the terms, we have:

8sinx*cosx = 8

sinx*cosx = 1

Since sinx*cosx = 1, this implies that either sinx = 1 and cosx = 1 or sinx = -1 and cosx = -1.

For sinx = 1 and cosx = 1, x can be any angle in the form x = 2nπ, where n is an integer.

For sinx = -1 and cosx = -1, x can be any angle in the form x = (2n + 1)π/2, where n is an integer.

Therefore, the solutions to the equation 7cos^x - 8sinx*cosx + sin^x = 0 are:

x = 2nπ or x = (2n + 1)π/2, where n is an integer.