Question

5 - 8cos^2 x = sin 2x

Answer 1

решение во вложении

за мазню извините

Answer 2

To solve the equation 5 - 8cos^2(x) = sin(2x), we can use trigonometric identities to simplify it and find the values of x that satisfy the equation.

Let's start by using the double angle identity for sine: sin(2x) = 2sin(x)cos(x).

The equation becomes:

5 - 8cos^2(x) = 2sin(x)cos(x)

Rearranging the terms, we have:

8cos^2(x) + 2sin(x)cos(x) - 5 = 0

Now, let's simplify further by using the identity: cos^2(x) + sin^2(x) = 1.

Substituting cos^2(x) = 1 - sin^2(x) into the equation, we get:

8(1 - sin^2(x)) + 2sin(x)(1 - sin^2(x)) - 5 = 0

Expanding and rearranging the terms:

8 - 8sin^2(x) + 2sin(x) - 2sin^3(x) - 5 = 0

Combining like terms:

-2sin^3(x) - 8sin^2(x) + 2sin(x) + 3 = 0

Now, let's factor out a sin(x) term:

sin(x)(-2sin^2(x) - 8sin(x) + 2) + 3 = 0

We can see that sin(x) = 0 is one solution, so x = 0° or x = π radians.

To find the remaining solutions, we need to solve the quadratic equation -2sin^2(x) - 8sin(x) + 2 = 0.

Dividing the equation by -2:

sin^2(x) + 4sin(x) - 1 = 0

Now, we can use the quadratic formula to solve for sin(x):

sin(x) = (-b ± √(b^2 - 4ac)) / (2a)

sin(x) = (-4 ± √(4^2 - 4(1)(-1))) / (2(1))

sin(x) = (-4 ± √(16 + 4)) / 2

sin(x) = (-4 ± √20) / 2

sin(x) = (-4 ± 2√5) / 2

sin(x) = -2 ± √5

Since the range of the sine function is -1 ≤ sin(x) ≤ 1, the solution -2 + √5 is not valid.

Therefore, the remaining valid solutions for sin(x) are:

sin(x) = -2 - √5

To find the corresponding values of x, we can use the inverse sine function:

x = arcsin(-2 - √5)

Using a calculator, we find:

x ≈ -1.855 radians or x ≈ -106.32°

Therefore, the solutions to the equation 5 - 8cos^2(x) = sin(2x) are:

x = 0° or x = π radians (corresponding to sin(x) = 0)

x ≈ -106.32° or x ≈ -1.855 radians (corresponding to sin(x) = -2 - √5)