Question

Sin 3x cos x - cos 3x sin x ≤ 1/2

Answer 1

sin на cos плюс cos на sin даст sin суммы. sin (3х+х)=0,5

4х= минус 1 в степени k умноженное на пи делить на 6 + пи*k/

х= минус 1 в степени k умноженное на пи делить на 24 и пи умнож. на k деленное на 4

Answer 2

Ответ:

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Answer 3

To prove the inequality $\sin(3x)\cos(x) - \cos(3x)\sin(x) \leq \frac{1}{2}$, we can use the angle sum and difference identities for trigonometric functions.

First, we can simplify the left side of the inequality using the identities:

1. $\sin(A - B) = \sin(A)\cos(B) - \cos(A)\sin(B)$
2. $\sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B)$

Let's apply these identities to your expression:

$\sin(3x)\cos(x) - \cos(3x)\sin(x) = \sin(3x - x) = \sin(2x)$

Now, we have the inequality $\sin(2x) \leq \frac{1}{2}$.

To find the values of $x$ that satisfy this inequality, we can consider the unit circle. The sine function represents the y-coordinate of points on the unit circle.

1. The maximum value of $\sin(2x)$ is 1, which occurs when $2x = \frac{\pi}{2} + 2n\pi$, where $n$ is an integer.
2. The minimum value of $\sin(2x)$ is -1, which occurs when $2x = -\frac{\pi}{2} + 2n\pi$, where $n$ is an integer.

Now, we want to find the values of $x$ such that $\sin(2x) \leq \frac{1}{2}$. We can use these inequalities:

1. $2x \leq \frac{\pi}{2} + 2n\pi$
2. $2x \geq -\frac{\pi}{2} + 2n\pi$

Dividing both sides by 2 in each inequality:

1. $x \leq \frac{\pi}{4} + n\pi$
2. $x \geq -\frac{\pi}{4} + n\pi$

Now, to find the values of $x$ that satisfy both inequalities simultaneously, you can choose integer values of $n$. These values will give you the range of $x$ for which $\sin(2x) \leq \frac{1}{2}$. The solutions will be:

$-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$

So, the values of $x$ in this range satisfy the inequality $\sin(2x) \leq \frac{1}{2}$.