Question

2cos2x+2sin^x-sin2x = 0

Answer 1

Смотри.....................

Answer 2

It seems like you have a trigonometric equation you'd like to solve:

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

To solve this equation, we can use some trigonometric identities to simplify it. First, let's use the double angle formulas for sine and cosine:

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

Now, substitute these identities back into the equation:

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

Now, simplify further:

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

Next, factor out common terms:

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

Now, we can try to solve for cos(x):

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

Divide the equation by 2 to simplify:

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

Now, let's focus on the term involving cos(x):

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

Use the identity sin(2x) = 2sin(x)cos(x):

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

Now, we can use the identity cos^2(x) = 1 - sin^2(x):

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

Rearrange the terms:

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

Now, we have a quadratic equation in sin(x). Let's multiply everything by 2 to get rid of fractions:

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

Combine like terms:

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

Now, this is a quadratic equation in sin(x). You can solve it using the quadratic formula:

sin(x) = [-b ± sqrt(b^2 - 4ac)] / (2a)

In this case, a = -2, b = -1, and c = 1. Plug these values into the formula to find sin(x):

sin(x) = [1 ± sqrt((-1)^2 - 4(-2)(1))] / (2(-2))

sin(x) = [1 ± sqrt(1 + 8)] / (-4)

sin(x) = [1 ± sqrt(9)] / (-4)

sin(x) = [1 ± 3] / (-4)

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

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

Now, you can find the corresponding values of x by taking the arcsin (inverse sine) of these values:

1. For sin(x) = 1/2: x = arcsin(1/2) = π/6 (or 30 degrees)
2. For sin(x) = -1: x = arcsin(-1) = -π/2 (or -90 degrees)

So, the solutions to the equation 2cos(2x) + 2sin(x) - sin(2x) = 0 are:

x = π/6 + 2nπ (where n is an integer) x = -π/2 + 2nπ (where n is an integer)

These are the general solutions to the equation.