Question

3sin^2+2sinx cosx-cos^2x=0

Answer 1

Здесь надо всего лишь знать формулы

Answer 2

The given equation is:

3sin^2(x) + 2sin(x)cos(x) - cos^2(x) = 0

To solve this equation, we can simplify it using trigonometric identities.

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

3(1 - cos^2(x)) + 2sin(x)cos(x) - cos^2(x) = 0

Expanding the equation:

3 - 3cos^2(x) + 2sin(x)cos(x) - cos^2(x) = 0

Rearranging the terms:

-4cos^2(x) + 2sin(x)cos(x) + 3 = 0

Now, let's factor out a common factor of 2:

2(-2cos^2(x) + sin(x)cos(x)) + 3 = 0

Next, we can simplify the expression inside the parentheses by using the identity sin(2x) = 2sin(x)cos(x):

2(-2cos^2(x) + 2sin(x)cos(x)) + 3 = 0

Simplifying further:

-4cos^2(x) + 4sin(x)cos(x) + 3 = 0

We can divide the entire equation by 4 to simplify it:

-cos^2(x) + sin(x)cos(x) + 3/4 = 0

Now, let's introduce a substitution. Let's say y = cos(x):

-y^2 + sin(x)y + 3/4 = 0

This equation is quadratic in terms of y. We can solve it using the quadratic formula:

y = [-sin(x) ± √(sin^2(x) - 4(-3/4))] / (-2)

Simplifying further:

y = [-sin(x) ± √(sin^2(x) + 3)] / (-2)

Now, substitute y = cos(x) back into the equation:

cos(x) = [-sin(x) ± √(sin^2(x) + 3)] / (-2)

At this point, it becomes difficult to find an exact solution because the equation involves both sine and cosine functions. However, you can use numerical methods or graphing techniques to approximate the solutions.