Question

9sinxcosx-7 cos2x=2sin^2x

Answer 1

Вроде так..

2sin²x-9sinxcosx+7cos²x=0 (:cos²x≠0)

2tg²x-9tgx+7=0

tgx=y;2y²-9y+7=0;y₁=1;y₂=7/2

1) tgx=1;x=π/4+πn,n∈Z

2) tgx=7/2;x=arctg(7/2)+πk,k∈Z.

Answer 2

To solve the equation 9sin(x)cos(x) - 7cos(2x) = 2sin^2(x), we can use trigonometric identities and algebraic manipulations to simplify and find the solutions.

Let's start by using the double-angle formula for cosine: cos(2x) = 1 - 2sin^2(x). Replacing cos(2x) in the equation, we have:

9sin(x)cos(x) - 7(1 - 2sin^2(x)) = 2sin^2(x)

Expanding the equation:

9sin(x)cos(x) - 7 + 14sin^2(x) = 2sin^2(x)

Now, let's gather the terms with sin^2(x) on one side of the equation:

14sin^2(x) - 2sin^2(x) - 9sin(x)cos(x) + 7 = 0

Combining like terms:

12sin^2(x) - 9sin(x)cos(x) + 7 = 0

We can notice that this equation is quadratic in sin(x). To solve it, we can use the quadratic formula:

sin(x) = [-b ± √(b^2 - 4ac)] / (2a)

For our equation, a = 12, b = -9cos(x), and c = 7. Substituting these values into the quadratic formula, we get:

sin(x) = [9cos(x) ± √((-9cos(x))^2 - 4 * 12 * 7)] / (2 * 12)

Simplifying:

sin(x) = [9cos(x) ± √(81cos^2(x) - 336)] / 24

Now, we can solve for sin(x) by considering the two possible signs:

1. sin(x) = [9cos(x) + √(81cos^2(x) - 336)] / 24
2. sin(x) = [9cos(x) - √(81cos^2(x) - 336)] / 24

Once we find the values of sin(x), we can use the trigonometric identity sin^2(x) + cos^2(x) = 1 to find the corresponding values of cos(x).