Question

Sin^2x+2sinx*cosx=0​

Answer 1

Ответ:

$\sin(2x) + 2 \sin(x) \times \cos(x) = 0 \\ \sin(2x) + \sin(2x) = 0 \\ 2 \sin(2x) = 0 \\ \sin(2x) = 0 \\ 2x = k\pi \\ x = \frac{k\pi}{2}$

Answer 2

To solve the equation sin^2(x) + 2sin(x)cos(x) = 0, we can use trigonometric identities to simplify it.

First, we notice that sin^2(x) = (sin(x))^2 and cos(x) = cos(x)^2 - sin(x)^2.

Substituting these identities into the original equation:

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

Next, distribute the 2sin(x) into the parentheses:

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

Now, let's try to factor out common terms:

sin(x) * [sin(x) + 2cos(x)^2 - 2sin(x)^2] = 0

Now, we have a product of two factors equal to zero. According to the zero-product property, for the entire equation to be true, one of the factors must be equal to zero:

1. sin(x) = 0

To solve for x in this case, we look for all angles within the range [0, 2π] where the sine function is zero. The solutions are:

x = 0, π

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

Now, we need to solve this equation for x. First, we'll move the sin(x) term to the other side:

2cos(x)^2 - 2sin(x)^2 = -sin(x)

Next, we'll use the Pythagorean identity sin^2(x) + cos^2(x) = 1 to replace sin(x)^2:

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

Now, simplify:

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

Combine like terms:

4cos^2(x) - 2 = -sin(x)

Rearrange the equation:

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

Now, we can use the Pythagorean identity again to replace cos^2(x):

sin(x) = -4(1 - sin^2(x)) + 2

Expand:

sin(x) = -4 + 4sin^2(x) + 2

Combine like terms:

sin(x) = 4sin^2(x) - 2

Move all terms to one side of the equation:

4sin^2(x) - sin(x) - 2 = 0

Now, we have a quadratic equation in terms of sin(x). To solve this, you can use the quadratic formula or factor it if possible. Unfortunately, this quadratic equation cannot be factored easily, so we'll use the quadratic formula:

sin(x) = [-b ± √(b^2 - 4ac)] / 2a

where a = 4, b = -1, and c = -2.

sin(x) = [1 ± √(1 - 4(4)(-2))] / 2(4)

sin(x) = [1 ± √(1 + 32)] / 8

sin(x) = [1 ± √33] / 8

Now, we have two possible solutions for sin(x):

a) sin(x) = (1 + √33) / 8 b) sin(x) = (1 - √33) / 8

To find the corresponding angles, use the inverse sine function (sin^-1):

a) x = sin^-1((1 + √33) / 8) b) x = sin^-1((1 - √33) / 8)

Note: Some of these solutions might involve complex numbers. Be sure to check the range of your angle if you are looking for real solutions within a specific range.