Question

4sin3x*cos3x>√2 sin²3х-соs²3x>-0.5​

Answer 1

Ответ:

Объяснение:

1) 2sin3xcos3x>√2/2

sin6x>√2/2

П/4+2Пk<6x<3П/4+2Пk

П/24+Пk/3<x<П/8+Пk/3

2)cos6x<1/2

П/3+2Пk<6x<11П/6+2Пk

П/18+Пk/3<x<11П/36+Пk/3

Answer 2

To solve the inequality $4\sin(3x)\cos(3x) > \sqrt{2}\sin^2(3x) - \cos^2(3x) > -0.5$, we'll break it down into two separate inequalities:

1. $4\sin(3x)\cos(3x) > \sqrt{2}\sin^2(3x) - \cos^2(3x)$
2. $\sqrt{2}\sin^2(3x) - \cos^2(3x) > -0.5$

Let's solve each of these inequalities separately:

Inequality 1: $4\sin(3x)\cos(3x) > \sqrt{2}\sin^2(3x) - \cos^2(3x)$

First, notice that $\sqrt{2}$ can be written as $2^{1/2}$, and $\sin^2(3x) = 1 - \cos^2(3x)$ (using the Pythagorean identity for sine and cosine). Substituting these values, we get:

$4\sin(3x)\cos(3x) > 2^{1/2} (1 - \cos^2(3x)) - \cos^2(3x)$

Simplify:

$4\sin(3x)\cos(3x) > 2^{1/2} - 2^{1/2}\cos^2(3x) - \cos^2(3x)$

Combine the cosine terms:

$4\sin(3x)\cos(3x) > 2^{1/2} - (2^{1/2} + 1)\cos^2(3x)$

Now, divide both sides by $4\cos(3x)$ (note that we assume $\cos(3x) \neq 0$):

$\sin(3x) > \frac{2^{1/2} - (2^{1/2} + 1)\cos^2(3x)}{4\cos(3x)}$

$\sin(3x) > \frac{2^{1/2}}{4\cos(3x)} - \frac{2^{1/2} + 1}{4}\cos(3x)$

This inequality holds for specific ranges of $x$ where $\sin(3x)$ and $\cos(3x)$ are positive or negative. However, solving this inequality analytically may be quite complex and may involve finding the exact intervals of $x$ where the inequality holds.

Inequality 2: $\sqrt{2}\sin^2(3x) - \cos^2(3x) > -0.5$

This inequality is relatively simpler to solve. First, move all terms to one side:

$\sqrt{2}\sin^2(3x) - \cos^2(3x) + 0.5 > 0$

Now, you can use substitution to replace $\sin^2(3x)$ using the identity $\sin^2(3x) = 1 - \cos^2(3x)$:

$\sqrt{2}(1 - \cos^2(3x)) - \cos^2(3x) + 0.5 > 0$

Distribute $\sqrt{2}$: