Question

Sin2x+2sin*в_квадрате*x=0

Answer 1

$sin2x+2sin^2x=0 \\ 2sinxcosx+2sin^2x=0 \\ 2sinx(cosx+sinx)=0 \\ \\ sinx=0 \\ x= \pi k , k \in z \\ \\ cosx+sinx=0 \\ cosx=-sinx \ (:cosx \neq 0) \\ tgx=-1 \\ x=- \frac{ \pi }{4} + \pi k , k \in Z$

Answer 2

To solve the equation sin(2x) + 2sin^2(x) = 0, we can use some trigonometric identities to simplify the equation and find the solutions. First, let's rewrite the equation using the double angle identity for sine:

sin(2x) = 2sin(x)cos(x)

Now, the equation becomes:

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

Next, we can factor out a common term of 2sin(x):

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

Now, to find the solutions, we set each factor equal to zero:

1. 2sin(x) = 0 sin(x) = 0 x = 0° + n*180° where n is an integer (e.g., n = 0, 1, -1, 2, -2, ...)

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

To solve the second equation, we'll use another trigonometric identity, namely, the Pythagorean identity:

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

Rearrange the equation to solve for cos(x):

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

Now substitute this into the equation cos(x) + sin(x) = 0:

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

Rearrange and solve for sin(x):

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

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

Now, we have a quadratic equation in terms of sin(x). Use the quadratic formula:

sin(x) = [ -(-1) ± √((-1)^2 - 41(-1)) ] / (2*1)

sin(x) = [ 1 ± √(1 + 4) ] / 2

sin(x) = [ 1 ± √5 ] / 2

So the two solutions for sin(x) are:

a) sin(x) = [1 + √5] / 2 b) sin(x) = [1 - √5] / 2

Now, to find the corresponding angles, we can use the inverse sine function:

a) x = sin^(-1)((1 + √5) / 2) b) x = sin^(-1)((1 - √5) / 2)

Please note that the angles obtained from the inverse sine function are limited to a certain range. To get all possible solutions, you can add multiples of 360° to the angles obtained above:

x = sin^(-1)((1 + √5) / 2) + n * 360° x = sin^(-1)((1 - √5) / 2) + n * 360°

where n is an integer (e.g., n = 0, 1, -1, 2, -2, ...). These are all the solutions to the given equation.