Question

Log2(cos(2x)+ cos(x/2)) + log1/2(sin(x) + cos(x/2)) = 0

Answer 1

решение на фото==================

Answer 2

To solve the equation

log2(cos(2x) + cos(x/2)) + log1/2(sin(x) + cos(x/2)) = 0,

we'll use logarithmic properties to simplify it. First, let's recall some logarithmic properties:

1. log_b(a) + log_b(c) = log_b(ac)
2. log_b(a) - log_b(c) = log_b(a/c)
3. log_b(a^n) = n * log_b(a)

Now, let's apply these properties to the given equation:

log2(cos(2x) + cos(x/2)) + log1/2(sin(x) + cos(x/2)) = 0

Using property 1 (combining the logarithms):

log2(cos(2x) + cos(x/2)) * 1/2 = 0

Now, to eliminate the logarithm, we can rewrite the equation as an exponential equation:

2^(log2(cos(2x) + cos(x/2)) * 1/2) = 2^0

Now, recall that 2^0 = 1, and (a^(m/n)) = (n√a)^m:

cos(2x) + cos(x/2) = 1

Now, let's solve for x:

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

We can use the trigonometric identity: cos(2x) = 2cos^2(x) - 1

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

Now, we have a quadratic equation in terms of cos(x):

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

To solve this quadratic equation, we can use the quadratic formula:

cos(x) = [-b ± √(b^2 - 4ac)] / 2a

In this case, a = 2, b = 1, and c = -2.

cos(x) = [-(1) ± √((1)^2 - 4(2)(-2))] / 2(2)

cos(x) = [-1 ± √(1 + 16)] / 4

cos(x) = [-1 ± √17] / 4

So we have two possible solutions for cos(x):

1. cos(x) = (-1 + √17) / 4
2. cos(x) = (-1 - √17) / 4

Now, to find the values of x, you can take the inverse cosine (cos^(-1)) of these solutions:

1. x = cos^(-1)((-1 + √17) / 4)
2. x = cos^(-1)((-1 - √17) / 4)

Keep in mind that these solutions are in terms of cosine, and you may need to use a calculator to approximate the values of x.