Question

Cos x / 1 + c o s 2 x < 0

Answer 1

Если дробь <0, точислитель и знаменатель разных знаков, то есть

$\frac{a}{b}\ \textless \ 0\; \; \to \; \; \left \{ {{a\ \textgreater \ 0} \atop {b\ \textless \ 0}} \right. \; \;ili\; \; \left \{ {{a\ \textless \ 0} \atop {b\ \textgreater \ 0}} \right.$

Но знаменатель < 0  не может быть т.к.тогда

$1+cos2x\ \textless \ 0\; \; \to \; \; cos2x\ \textless \ -1.$

А тогда cos2x  может быть только = -1  ( $-1 \leq cos2x \leq 1$ ). Но тогда хнаменатель обратиться в 0, чего не должно быть.

$\left \{ {{cosx\ \textless \ 0} \atop {cos2x\ \textgreater \ 0}} \right. \; \left \{ {{\frac{\pi}{2}+2\pi n \leq x \leq \frac{3\pi}{2}+2\pi n} \atop {-\frac{\pi}{2}+2\pi k \leq 2x \leq \frac{\pi}{2}+2\pi k}} \right. \; \left \{ {{\frac{\pi}{2}+2\pi n \leq x \leq \frac{3\pi}{2}+2\pi n} \atop {-\frac{\pi}{4}+\pi k \leq x \leq \frac{\pi}{4}+\pi k}} \right. \; \to \\\\\frac{3\pi}{4}+2\pi k \leq x \leq \frac{5\pi}{4}+2\pi k,\; k\in Z$

Answer 2

To find the values of $x$ for which $\frac{\cos(x)}{1 + \cos^2(x)} < 0$, you need to consider the signs of $\cos(x)$ and $1 + \cos^2(x)$.

First, let's find the values of $x$ for which $\cos(x) = 0$. The cosine function is equal to zero at certain angles, namely $x = \frac{\pi}{2} + k\pi$, where $k$ is an integer.

Next, let's find the values of $x$ for which $1 + \cos^2(x) < 0$. However, this inequality can never be satisfied because $\cos^2(x)$ is always non-negative, and when you add 1 to it, the result is always positive or zero. So, there are no values of $x$ for which $1 + \cos^2(x) < 0$.

Now, consider the ratio $\frac{\cos(x)}{1 + \cos^2(x)}$. Since the denominator is always positive or zero and the numerator $\cos(x)$ can be either positive, negative, or zero depending on the value of $x$, the sign of the ratio will depend on the sign of $\cos(x)$.

1. When $\cos(x) > 0$, the ratio is positive.
2. When $\cos(x) < 0$, the ratio is negative.
3. When $\cos(x) = 0$, the ratio is undefined (division by zero).

Therefore, the inequality $\frac{\cos(x)}{1 + \cos^2(x)} < 0$ is satisfied when $\cos(x) < 0$.

To summarize, the solution to the inequality is $x$ belonging to the set of angles where $\cos(x) < 0$, which corresponds to the intervals:

$$x \in \left(\frac{\pi}{2} + 2k\pi, \frac{3\pi}{2} + 2k\pi\right), \text{ where } k \text{ is an integer}.$$

These intervals represent the values of $x$ for which the given inequality holds true.