Question

12sin 2x+sin x +cos x+6=0

Answer 1

Пошаговое объяснение:

это весь ответ, как смогла так и помогла,

Answer 2

To solve the equation 12sin(2x) + sin(x) + cos(x) + 6 = 0, we can start by using trigonometric identities to simplify the equation. Let's break it down step by step:

1. We can use the double angle formula for sine: sin(2x) = 2sin(x)cos(x). Rewriting the equation with this substitution, we get: 12(2sin(x)cos(x)) + sin(x) + cos(x) + 6 = 0.

2. Rearranging the terms, we have: 24sin(x)cos(x) + sin(x) + cos(x) + 6 = 0.

3. Let's combine the sin(x) and cos(x) terms: sin(x)(24cos(x) + 1) + cos(x) + 6 = 0.

4. Now, let's focus on the parentheses (24cos(x) + 1): We can square both sides of the equation to eliminate the square root: (24cos(x) + 1)^2 = 1. Expanding the square, we get: 576cos^2(x) + 48cos(x) + 1 = 1.

5. Simplifying the equation further, we have: 576cos^2(x) + 48cos(x) = 0.

6. Factoring out common terms, we get: 48cos(x)(12cos(x) + 1) = 0.

Now we have two cases to consider:

Case 1: 48cos(x) = 0. This implies cos(x) = 0. Solving for x, we have: x = π/2 + nπ, where n is an integer.

Case 2: 12cos(x) + 1 = 0. Subtracting 1 from both sides, we get: 12cos(x) = -1. Dividing both sides by 12, we have: cos(x) = -1/12. Taking the inverse cosine (arccos) of both sides: x = arccos(-1/12) + 2πn or x = -arccos(-1/12) + 2πn, where n is an integer.

So the solutions to the equation 12sin(2x) + sin(x) + cos(x) + 6 = 0 are: x = π/2 + nπ, where n is an integer, or x = arccos(-1/12) + 2πn or x = -arccos(-1/12) + 2πn, where n is an integer.