Question

2sin²x+3sinxcosx=2cos²x​

Answer 1

Ответ:

x= -1,10715 + k n 0,463648 + k n , k Э Z

Объяснение:

ezi

Answer 2

To solve the trigonometric equation 2sin²x + 3sin(x)cos(x) = 2cos²x, we can use trigonometric identities to simplify and rearrange the equation. Let's start by applying some trigonometric identities:

1. sin(2x) = 2sin(x)cos(x)
2. 1 - cos²x = sin²x (from the Pythagorean identity)

Now, let's rewrite the equation using these identities:

2sin²x + 3sin(x)cos(x) = 2cos²x

2sin²x + 3sin(x)cos(x) = 2(1 - sin²x)

Now, we can solve for sin(x):

2sin²x + 3sin(x)cos(x) = 2 - 2sin²x

Combine like terms:

4sin²x + 3sin(x)cos(x) - 2 = 0

Now, let's use a substitution to simplify the equation. Let u = sin(x):

4u² + 3ucos(x) - 2 = 0

Now, this is a quadratic equation in terms of u. We can solve it using the quadratic formula:

u = (-B ± √(B² - 4AC)) / (2A)

In this case, A = 4, B = 3cos(x), and C = -2. Substituting these values:

u = (-3cos(x) ± √((3cos(x))² - 4(4)(-2))) / (2(4))

Simplify further:

u = (-3cos(x) ± √(9cos²(x) + 32)) / 8

Now, we have two possibilities for u:

1. u = (-3cos(x) + √(9cos²(x) + 32)) / 8
2. u = (-3cos(x) - √(9cos²(x) + 32)) / 8

Since u = sin(x), and sin(x) cannot exceed the range [-1, 1], we need to find the values of cos(x) that satisfy these conditions. Let's solve for cos(x) in both cases:

1. (-1) ≤ -3cos(x) + √(9cos²(x) + 32) / 8 ≤ 1
2. (-1) ≤ -3cos(x) - √(9cos²(x) + 32) / 8 ≤ 1

Now, you can solve each of these inequalities for cos(x). The solutions for cos(x) that satisfy these inequalities will be the solutions to the original trigonometric equation 2sin²x + 3sin(x)cos(x) = 2cos²x. Please note that this might involve solving a system of inequalities, and the exact solutions may vary depending on the values of u that satisfy the inequalities.