Question

Log_11(√2sinx)-1/2log_11(cosx+2)=0

Answer 1

Ответ прилагается на фото, там вроде понятно .

Пошаговое объяснение: Извини первое и второе вложение перепутанно. Второе - первое, первое- второе.

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Answer 2

To solve the equation log_11(√2sinx) - 1/2log_11(cosx + 2) = 0, follow these steps:

Step 1: Combine the logarithms using logarithmic properties. Using the rule log_a(b) - log_a(c) = log_a(b/c), you can combine the two logarithms on the left side:

log_11(√2sinx) - 1/2log_11(cosx + 2) = log_11(√2sinx / √(cosx + 2))

Step 2: Eliminate the square roots. Using the properties of exponents (a^(1/2) = √a), you can simplify the expression inside the logarithm:

√2sinx / √(cosx + 2) = √(2sinx / (cosx + 2))

Step 3: Solve for x. Now that you have a single logarithmic expression, you can set it equal to 0 and solve for x:

log_11(√(2sinx / (cosx + 2))) = 0

Using the property that log_a(b) = 0 if and only if b = 1, you can solve for the expression inside the logarithm:

√(2sinx / (cosx + 2)) = 1

Square both sides to get rid of the square root:

2sinx / (cosx + 2) = 1

Multiply both sides by (cosx + 2):

2sinx = cosx + 2

Now you have a trigonometric equation. To solve for x, rearrange the equation:

2sinx - cosx = 2

Now you can use trigonometric identities to simplify the left side of the equation. One way to do this is to use the identity sin^2x + cos^2x = 1:

2(sin^2x + cos^2x) - cosx = 2

2sin^2x + 2cos^2x - cosx = 2

Substitute 1 - sin^2x for cos^2x (from the identity):

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

2sin^2x + 2 - 2sin^2x - cosx = 2

2 - cosx = 2

Now, isolate the cosine term:

-cosx = 0

cosx = 0

This means that x must be an angle for which the cosine is 0, which occurs at π/2 radians and 3π/2 radians, and their multiples. So the solutions for x are:

x = π/2 + 2πn and x = 3π/2 + 2πn, where n is an integer.

These are the values of x that satisfy the given equation.