Question

Sin 2x* tgx +1 = 3 sinx​

Answer 1

Ответ:

решение и ответ на фото

Answer 2

To solve the equation sin(2x) * tan(x) + 1 = 3sin(x), we can apply some trigonometric identities and algebraic manipulations.

First, let's simplify the equation using the double angle formula for sine (sin(2x) = 2sin(x)cos(x)) and the identity for tangent (tan(x) = sin(x) / cos(x)):

2sin(x)cos(x) * (sin(x) / cos(x)) + 1 = 3sin(x)

Next, cancel out the common factors of cos(x):

2sin(x) * sin(x) + 1 = 3sin(x)

Simplify further:

2sin^2(x) + 1 = 3sin(x)

Rearrange the terms to form a quadratic equation:

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

Now, let's solve this quadratic equation. We can use factoring, completing the square, or the quadratic formula. In this case, factoring might not be straightforward, so let's use the quadratic formula:

sin(x) = (-b ± √(b^2 - 4ac)) / (2a)

Here, a = 2, b = -3, and c = 1. Substituting these values:

sin(x) = (-(-3) ± √((-3)^2 - 4 * 2 * 1)) / (2 * 2) sin(x) = (3 ± √(9 - 8)) / 4 sin(x) = (3 ± √1) / 4 sin(x) = (3 ± 1) / 4

This gives us two possibilities:

1. sin(x) = (3 + 1) / 4 = 4 / 4 = 1
2. sin(x) = (3 - 1) / 4 = 2 / 4 = 1/2

Let's solve for x in each case:

1. sin(x) = 1

Since sin(x) equals 1, this occurs when x = π/2 + 2πn, where n is an integer.

2. sin(x) = 1/2

This occurs when x = π/6 + 2πn or x = 5π/6 + 2πn, where n is an integer.

Therefore, the solutions to the equation sin(2x) * tan(x) + 1 = 3sin(x) are:

x = π/2 + 2πn, π/6 + 2πn, or 5π/6 + 2πn, where n is an integer.