Question

Cosx/ 1+cos2x меньше 0

Answer 1

Cosx/(1+cos2x)<0 => cosx/(sin(^2)x+cos(^2)x+cos(^2)x-sin(^2)x)<0 => cosx/2cos(^2)x<0 => 1/2cosx<0 => cosx>1/2 => x€(-pi/3+2pi*k; pi/3+2pi*k), k€Z.

Answer 2

To determine when the expression $\frac{\cos(x)}{1+\cos^2(x)}$ is less than 0, we need to find the intervals where the expression is negative.

First, let's find the critical points where the expression is equal to 0 or undefined. The expression is undefined when the denominator $1 + \cos^2(x) = 0$. However, this never happens since $\cos^2(x)$ is always non-negative (between 0 and 1), and thus $1 + \cos^2(x)$ is always positive.

Now, let's look at the sign of $\cos(x)$. The cosine function is positive in the first and fourth quadrants (0 to $\frac{\pi}{2}$ and $2\pi$ to $\frac{3\pi}{2}$) and negative in the second and third quadrants ($\frac{\pi}{2}$ to $\pi$ and $\frac{3\pi}{2}$ to $2\pi$).

Next, let's analyze the sign of the denominator $1 + \cos^2(x)$. Since $\cos^2(x)$ is always non-negative, $1 + \cos^2(x)$ is always positive, and hence, it doesn't affect the overall sign of the expression.

Now, we need to combine the signs of the numerator and denominator. If the numerator and denominator have the same sign, the fraction is positive. If they have opposite signs, the fraction is negative.

Let's summarize:

1. When $\cos(x) > 0$ (first and fourth quadrants): $\frac{\cos(x)}{1+\cos^2(x)} > 0$ since both the numerator and denominator are positive.

2. When $\cos(x) < 0$ (second and third quadrants): $\frac{\cos(x)}{1+\cos^2(x)} < 0$ since the numerator is negative, and the denominator is positive.

So, the expression $\frac{\cos(x)}{1+\cos^2(x)}$ is less than 0 in the second and third quadrants of the unit circle ($\frac{\pi}{2}$ to $\pi$ and $\frac{3\pi}{2}$ to $2\pi$).