Question

3sin^2x+11sin x*cos x+2 cos^2 x=-3

Answer 1

Решение смотри на фотографии

Answer 2

To solve the trigonometric equation 3sin^2x + 11sin x*cos x + 2cos^2x = -3, we can use some trigonometric identities and algebraic manipulations to simplify it and find the solutions.

First, we will use the Pythagorean identity: sin^2x + cos^2x = 1.

Now, let's rewrite the equation using the Pythagorean identity:

3(sin^2x + cos^2x) + 11sin x*cos x + 2cos^2x = -3

Next, distribute the 3:

3sin^2x + 3cos^2x + 11sin x*cos x + 2cos^2x = -3

Combine like terms:

3sin^2x + 5cos^2x + 11sin x*cos x = -3

Now, replace sin^2x with (1 - cos^2x) using the Pythagorean identity:

3(1 - cos^2x) + 5cos^2x + 11sin x*cos x = -3

Distribute the 3:

3 - 3cos^2x + 5cos^2x + 11sin x*cos x = -3

Simplify further:

-3cos^2x + 11sin x*cos x = -6

Divide the entire equation by -1:

3cos^2x - 11sin x*cos x = 6

Now, we have a quadratic-like expression in cos x. Let's make a substitution: Let u = cos x.

The equation becomes:

3u^2 - 11sin x*u - 6 = 0

Now, we can solve this quadratic equation for u using the quadratic formula:

u = [ -(-11sin x) ± √((-11sin x)^2 - 4 * 3 * (-6)) ] / 2 * 3

Simplifying further:

u = [ 11sin x ± √(121sin^2 x + 72) ] / 6

Since u = cos x:

cos x = [ 11sin x ± √(121sin^2 x + 72) ] / 6

Now, we have two possible solutions for cos x. To find the solutions for sin x, we can use the Pythagorean identity again:

sin^2x + cos^2x = 1

sin^2x + ([ 11sin x ± √(121sin^2 x + 72) ] / 6)^2 = 1

Now, we have two separate equations for sin x:

1. sin^2x + ([ 11sin x + √(121sin^2 x + 72) ] / 6)^2 = 1

2. sin^2x + ([ 11sin x - √(121sin^2 x + 72) ] / 6)^2 = 1

We can solve these two equations separately to find the values of sin x. Once we have the values of sin x, we can find the corresponding values of cos x using the equation cos x = [ 11sin x ± √(121sin^2 x + 72) ] / 6.