Question

8sin2x - 16cosx = 0 Помогите пожалуйста.

Answer 1

Вспомним  sin2x=  2sinx.cosx  формуле  и   получим  16 sinx.  cosx-16cosx =0.  16cosx (sinx-1)  =0  . cosx =0 или     sinx=1 .cosx=0  x=90+180k  .   sinx=1   x=90+360k . Так  как  второй   решение   принадлежит    в  первом   решение  .то   получим  x=90+ 180 *k  .  

Answer 2

Solving the Equation 8sin(2x) - 16cos(x) = 0

To solve the equation 8sin(2x) - 16cos(x) = 0, we can use trigonometric identities to simplify and solve for the value of x.

First, let's express the sine and cosine functions in terms of a single trigonometric function using the double-angle identity for sine and the Pythagorean identity for cosine.

The double-angle identity for sine is: sin(2x) = 2sin(x)cos(x)

The Pythagorean identity for cosine is: cos^2(x) + sin^2(x) = 1

We can use these identities to simplify the given equation.

Applying Trigonometric Identities

Using the double-angle identity for sine, we can rewrite the equation as: 8(2sin(x)cos(x)) - 16cos(x) = 0

Factoring out 8 and rearranging, we get: 16sin(x)cos(x) - 16cos(x) = 0

Now, we can factor out 16cos(x) from the terms: 16cos(x)(sin(x) - 1) = 0

Finding Solutions

From the factored form, we can see that the equation has two possible solutions: 1. 16cos(x) = 0 2. sin(x) - 1 = 0

Let's solve for x in each case.

Solution 1: 16cos(x) = 0

For the first case, when 16cos(x) = 0, we can solve for x by finding the values of x where cosine is equal to 0. This occurs at x = π/2 + nπ, where n is an integer.

Solution 2: sin(x) - 1 = 0

For the second case, when sin(x) - 1 = 0, we can solve for x by finding the values of x where sine is equal to 1. This occurs at x = π/2 + 2nπ, where n is an integer.

Conclusion

Therefore, the solutions to the equation 8sin(2x) - 16cos(x) = 0 are: - x = π/2 + nπ, where n is an integer - x = π/2 + 2nπ, where n is an integer