Question

Sin^{2}3x-cosx^{2}\geq-0.5

Answer 1

Sin^{2}3x-cosx^{2}\geq-0.5

Ответ:

Х принадлежит R (0;_0.5)

Answer 2

To solve the inequality $\sin^2(3x) - \cos^2(x) \geq -0.5$, we can break it down step by step.

First, let's simplify the left side of the inequality using trigonometric identities:

$\sin^2(3x) - \cos^2(x) = (1 - \cos^2(3x)) - \cos^2(x)$

Using the trigonometric identity $\sin^2(\theta) = 1 - \cos^2(\theta)$, we can simplify the expression further:

$\sin^2(3x) - \cos^2(x) = \sin^2(3x) - (1 - \sin^2(x))$ $= \sin^2(3x) - 1 + \sin^2(x)$

Now, the inequality becomes:

$\sin^2(3x) - 1 + \sin^2(x) \geq -0.5$

Let's combine the terms involving sine:

$2\sin^2(x) + \sin^2(3x) - 1 \geq -0.5$

We can see that $\sin^2(x)$ is always between 0 and 1, so $2\sin^2(x)$ is always greater than or equal to 0.

Now, let's focus on $\sin^2(3x)$. This term varies between 0 and 1 as well.

To make progress, we can see that the maximum value of $\sin^2(3x)$ is 1 (when $\sin(3x) = 1$), and the minimum value is 0. Therefore, $\sin^2(3x)$ will always be between 0 and 1.

Putting everything together:

$2\sin^2(x) + \sin^2(3x) - 1 \geq -0.5$

Since $2\sin^2(x)$ and $\sin^2(3x)$ are always non-negative, the smallest possible value of the left side is $0 + 0 - 1 = -1$. Since -1 is less than -0.5, the inequality is true for all values of $x$.

In conclusion, the inequality $\sin^2(3x) - \cos^2(x) \geq -0.5$ is true for all real values of $x$.